{"id":3362,"date":"2024-08-30T12:10:23","date_gmt":"2024-08-30T06:10:23","guid":{"rendered":"https:\/\/stage-wp.10minuteschool.com\/?p=3362"},"modified":"2025-01-26T16:13:42","modified_gmt":"2025-01-26T10:13:42","slug":"ideal-integrals","status":"publish","type":"post","link":"https:\/\/10minuteschool.com\/content\/ideal-integrals\/","title":{"rendered":"\u0986\u09a6\u09b0\u09cd\u09b6 \u0995\u09bf\u099b\u09c1 \u09af\u09cb\u0997\u099c \u09a8\u09bf\u09b0\u09cd\u09a3\u09df"},"content":{"rendered":"

(Determining a few Ideal Integrals)<\/b><\/span><\/h2>\n

\u0995\u09df\u09c7\u0995\u099f\u09bf \u0986\u09a6\u09b0\u09cd\u09b6 \u09af\u09cb\u0997\u099c<\/a><\/span> \u09a8\u09bf\u09b0\u09cd\u09a3\u09df:<\/span><\/h3>\n

A.\u00a0 \\int \\frac{d x}{a^{2}+x^{2}}=\\frac{1}{a} \\tan ^{-1} \\frac{x}{a}+c<\/span><\/p>\n

B.\u00a0 \\int \\frac{d x}{a^{2}-x^{2}}=\\frac{1}{2 a} \\ln \\left|\\frac{a+x}{a-x}\\right|+c, x \\neq \\pm a<\/span><\/p>\n

C.\u00a0 \\int \\frac{d x}{x^{2}-a^{2}}=\\frac{1}{2 a} \\ln \\left|\\frac{x-a}{x+a}\\right|+c, x \\neq \\pm a<\/span><\/p>\n

D. \\int \\frac{d x}{\\sqrt{a^{2}-x^{2}}}=\\sin ^{-1} \\frac{x}{a}+c<\/span><\/p>\n

E. \\int \\frac{d x}{\\sqrt{a^{2}+x^{2}}}=\\ln \\left|\\sqrt{a^{2}+x^{2}}+x\\right|+c<\/span><\/p>\n

F. \\int \\frac{d x}{\\sqrt{x^{2}-a^{2}}}=\\ln \\left|\\sqrt{x^{2}-a^{2}}+x\\right|+c<\/span><\/p>\n

G. \\int \\sqrt{a^{2}-x^{2}} d x=\\frac{x \\sqrt{a^{2}-x^{2}}}{2}+\\frac{a^{2}}{2} \\sin ^{-1} \\frac{x}{a}+c<\/span><\/p>\n

 <\/p>\n

\u09aa\u09cd\u09b0\u09ae\u09be\u09a3 (Proof):<\/b><\/p>\n

A.\u00a0 \\int \\frac{d x}{a^{2}+x^{2}}=\\int \\frac{d x}{a^{2}\\left\\{1+\\left(\\frac{x}{a}\\right)^{2}\\right\\}}=\\frac{1}{a^{2}} \\int \\frac{a\\left(\\frac{1}{a} d x\\right)}{1+\\left(\\frac{x}{a}\\right)^{2}}\\left[\\because d\\left(\\frac{x}{a}\\right)=\\frac{1}{a} d x\\right]<\/span><\/p>\n=\\frac{1}{a} \\tan ^{-1} \\frac{x}{a}+c \\quad\\left[\\because \\int \\frac{d x}{1+x^{2}}=\\tan ^{-1} x+c\\right]<\/span>\n

\u00a0 \u00a0 \u00a0 \u00a0<\/span><\/p>\n

 <\/p>\n

\\text { B. } \\int \\frac{d x}{a^{2}-x^{2}}=\\int \\frac{d x}{(a-x)(a+x)}<\/span><\/span><\/p>\n

=\\int \\frac{1}{2 a}\\left\\{\\frac{1}{a+x}+\\frac{1}{a-x}\\right\\} d x=\\frac{1}{2 a}\\left\\{\\int \\frac{d x}{a+x}+\\int \\frac{-(-d x)}{a-x}\\right\\}<\/span><\/span><\/p>\n=\\frac{1}{2 a}\\{\\ln |x+a|-\\ln |x-a|\\}+c<\/span>\n

 <\/p>\n

=\\frac{1}{2 a} \\ln \\left|\\frac{a+x}{a-x}\\right|+c<\/span><\/span><\/p>\n

 <\/p>\n

C. \\int \\frac{d x}{x^{2}-a^{2}}=\\int \\frac{d x}{(x-a)(x+a)}<\/span><\/span><\/p>\n

=\\int \\frac{1}{2 a}\\left\\{\\frac{1}{x-a}-\\frac{1}{x+a}\\right\\} d x=\\frac{1}{2 a}\\left\\{\\int \\frac{d(x-a)}{x-a}-\\int \\frac{d x(x+a)}{x+a}\\right\\}<\/span><\/span><\/p>\n= \\frac{1}{2 a}\\{\\ln |x-a|-\\ln |x+a|\\}+c<\/span>\n

 <\/p>\n

=\\frac{1}{2 a} \\ln \\left|\\frac{x-a}{x+a}\\right|+c<\/span><\/span><\/p>\n

\u00a0 \u00a0 \u00a0<\/span><\/p>\n

D. \\int \\frac{d x}{\\sqrt{a^{2}-x^{2}}}=\\int \\frac{d x}{\\sqrt{a^{2}\\left\\{1-\\left(\\frac{x}{a}\\right)^{2}\\right.}}=\\int \\frac{a \\cdot\\left(\\frac{1}{a} d x\\right)}{a \\sqrt{1-\\left(\\frac{x}{a}\\right)^{2}}}=\\sin ^{-1} \\frac{x}{a}+c<\/span><\/span><\/p>\n

 <\/p>\n

\"\u09a8\u09bf\u09b0\u09cd\u09a3\u09df\"<\/p>\n

 <\/p>\n

\u00a0E. \u09ae\u09a8\u09c7 \u0995\u09b0\u09bf, x=a \\tan \\theta \\quad \\therefore d x=\\operatorname{asec}^{2} \\theta d \\theta<\/span><\/span><\/p>\n

\\therefore \\int \\frac{d x}{\\sqrt{a^{2}+x^{2}}}=\\int \\frac{\\operatorname{asec}^{2} \\theta d \\theta}{\\sqrt{a^{2}\\left(1+\\tan ^{2} \\theta\\right)}}<\/span><\/span><\/p>\n

=\\int \\frac{\\operatorname{asec}^{2} \\theta d \\theta}{\\sqrt{a^{2} \\sec ^{2} \\theta}}=\\int \\sec \\theta d \\theta<\/span><\/span><\/p>\n

=\\ln |\\sec \\theta+\\tan \\theta|+c_{1}<\/span><\/span><\/p>\n

=\\ln \\mid \\sqrt{1+\\frac{x^{2}}{a^{2}}+\\frac{x}{a}} \\mid+c_{1}<\/span><\/span><\/p>\n

=\\ln \\left|\\frac{\\sqrt{a^{2}+x^{2}}+x}{a}\\right|+c_{1}<\/span><\/span><\/p>\n

=\\ln \\left|\\sqrt{a^{2}+x^{2}}+x\\right|-\\ln |a|+c_{1}<\/span><\/span><\/p>\n

=\\ln \\left|\\sqrt{x^{2}+a^{2}}+x\\right|+c<\/span><\/span><\/p>\n

 <\/p>\n

F. \u09ae\u09a8\u09c7 \u0995\u09b0\u09bf, x=a \\sec \\theta \\quad \\therefore d x=a \\sec \\theta \\tan \\theta d \\theta<\/span><\/span><\/p>\n

\\therefore \\int \\frac{d x}{\\sqrt{x^{2}-a^{2}}}=\\int \\frac{\\operatorname{asec} \\theta \\tan \\theta d \\theta}{\\sqrt{a^{2}\\left(\\sec ^{2} \\theta-1\\right)}}<\/span><\/span><\/p>\n

=\\int \\frac{\\sec \\theta \\tan \\theta d \\theta}{\\tan \\theta}=\\int \\sec \\theta d \\theta<\/span><\/span><\/p>\n

=\\ln |\\sec \\theta+\\tan \\theta|+c_{1}<\/span><\/span><\/p>\n

=\\ln \\left|\\sec \\theta+\\sqrt{\\sec ^{2} \\theta-1}\\right|+c_{1}<\/span><\/span><\/p>\n

=\\ln \\left|\\frac{x}{a}+\\sqrt{\\frac{x^{2}}{a^{2}}-1}\\right|+c_{1}<\/span><\/span><\/p>\n

\\ln \\left|\\sqrt{x^{2}-a^{2}}+x\\right|+c<\/span><\/span><\/p>\n

 <\/p>\n

G. \u09ae\u09a8\u09c7 \u0995\u09b0\u09bf,, x=a \\sin \\theta \\Rightarrow \\theta=\\sin ^{-1} \\frac{x}{a} \\text { \u098f\u09ac\u0982 } d x=a \\cos \\theta d \\theta<\/span><\/span><\/p>\n

\\therefore \\int \\sqrt{a^{2}-x^{2}} d x=\\int \\sqrt{a^{2}\\left(1-\\sin ^{2} \\theta\\right)} a \\cos \\theta d \\theta<\/span><\/span><\/p>\n

=\\int a \\sqrt{\\left(1-\\sin ^{2} \\theta\\right)} a \\cos \\theta d \\theta<\/span><\/span><\/p>\n

=a^{2} \\int \\cos ^{2} \\theta d \\theta=a^{2} \\int \\frac{1}{2}(1+\\cos 2 \\theta) d \\theta<\/span><\/span><\/p>\n

=\\frac{a^{2}}{2}\\left(\\theta+\\frac{\\sin 2 \\theta}{2}\\right)+c<\/span><\/span><\/p>\n

=\\frac{a^{2}}{2}\\left(\\theta+\\frac{2 \\sin \\theta \\cos \\theta}{2}\\right)+c<\/span><\/span><\/p>\n

=\\frac{a^{2}}{2}\\left(\\theta+\\sin \\theta \\sqrt{1-\\sin ^{2} \\theta}\\right)+c<\/span><\/span><\/p>\n

=\\frac{a^{2}}{2}\\left(\\sin ^{-1} \\frac{x}{a}+\\frac{x}{a} \\sqrt{1-\\frac{x^{2}}{a^{2}}}\\right)+c<\/span><\/span><\/p>\n

=\\frac{a^{2}}{2} \\sin ^{-1} \\frac{x}{a}+\\frac{a^{2}}{2} \\cdot \\frac{x}{a^{2}} \\sqrt{a^{2}-x^{2}}+c<\/span><\/span><\/p>\n

=\\frac{x \\sqrt{a^{2}-x^{2}}}{2}+\\frac{a^{2}}{2} \\sin ^{-1} \\frac{x}{a}+c<\/span><\/span><\/p>\n

 <\/p>\n

\u09a6\u09cd\u09b0\u09b7\u09cd\u099f\u09ac\u09cd\u09af (Note):<\/b><\/p>\n

(a) \u09af\u09a6\u09bf \u09af\u09cb\u099c\u09cd\u09af \u09b0\u09be\u09b6\u09bf f(x),\\left(a^{2}-x^{2}\\right)^{1 \/ 2},\\left(a^{2}-x^{2}\\right)^{3 \/ 2}<\/span><\/span> \u0987\u09a4\u09cd\u09af\u09be\u09a6\u09bf \u0986\u0995\u09be\u09b0\u09c7 \u09a5\u09be\u0995\u09c7 \u09a4\u09ac\u09c7, <\/span>x = a sin\u03b8<\/span> \u09a7\u09b0\u09a4\u09c7 \u09b9\u09df\u0964<\/span><\/p>\n

(b) \u09af\u09a6\u09bf \u09af\u09cb\u099c\u09cd\u09af \u09b0\u09be\u09b6\u09bf f(x),\\left(a^{2}+x^{2}\\right)^{1 \/ 2},\\left(a^{2}+x^{2}\\right)^{3 \/ 2}<\/span> \u0987\u09a4\u09cd\u09af\u09be\u09a6\u09bf <\/span>\u0986\u0995\u09be\u09b0\u09c7 \u09a5\u09be\u0995\u09c7 \u09a4\u09ac\u09c7, <\/span>x = a tan\u03b8<\/span> \u09a7\u09b0\u09a4\u09c7 \u09b9\u09df\u0964<\/span><\/p>\n

(c) \u09af\u09a6\u09bf \u09af\u09cb\u099c\u09cd\u09af \u09b0\u09be\u09b6\u09bf f(x),\\left(x^{2}-a^{2}\\right)^{1 \/ 2},\\left(x^{2}-a^{2}\\right)^{3 \/ 2}<\/span><\/span>\u00a0\u0987\u09a4\u09cd\u09af\u09be\u09a6\u09bf \u0986\u0995\u09be\u09b0\u09c7 \u09a5\u09be\u0995\u09c7 \u09a4\u09ac\u09c7, <\/span>x = a sec\u03b8<\/span> \u09a7\u09b0\u09a4\u09c7 \u09b9\u09df\u0964<\/span><\/p>\n

 <\/p>\n

\u0989\u09a6\u09be\u09b9\u09b0\u09a3-\u09eb: \u09a8\u09bf\u099a\u09c7\u09b0 \u09af\u09cb\u0997\u099c\u0997\u09c1\u09b2\u09bf\u09b0 \u09ae\u09be\u09a8 \u09a8\u09bf\u09b0\u09cd\u09a3\u09df \u0995\u09b0:<\/b><\/p>\n

Example \u2013 5: Determine the following integrals<\/b><\/p>\n

(a)<\/span> \\int \\frac{d x}{9 x^{2}+4}<\/span><\/p>\n

(b)<\/span> \\int \\frac{d x}{\\sqrt{9-16 x^{2}}}<\/span><\/p>\n

\u09b8\u09ae\u09be\u09a7\u09be\u09a8 (Solution):<\/b><\/p>\n

(a)<\/span> \\int \\frac{d x}{9 x^{2}+4}=\\int \\frac{\\frac{1}{3} \\cdot(3 d x)}{2^{2}+(3 x)^{2}}<\/span><\/p>\n=\\frac{1}{3} \\cdot \\frac{1}{2} \\tan ^{-1} \\frac{3 x}{2}+c<\/span>\n

 <\/p>\n

=\\frac{1}{6} \\tan ^{-1} \\frac{3 x}{2}+c<\/span>\u00a0 \u00a0 \u00a0(Ans)<\/b><\/p>\n

 <\/p>\n

(b) \\int \\frac{d x}{\\sqrt{9-16 x^{2}}}=\\int \\frac{d x}{\\sqrt{3^{2}-(4 x)^{2}}}<\/span><\/span><\/p>\n

=\\frac{1}{4} \\int \\frac{(4 d x)}{\\sqrt{3^{2}-(4 x)^{2}}}=\\frac{1}{4} \\sin ^{-1} \\frac{4 x}{3}+c<\/span>\u00a0 \u00a0(Ans)<\/b><\/p>\n

 <\/p>\n

\u09a7\u09b0\u09a8-\u09e9 (Type \u2013 3):<\/b><\/p>\n

\\int \\frac{p x+q}{a x^{2}+b x+c} d x, \\int \\frac{p x+q}{\\sqrt{a x^{2}+b x+c}} d x, \\int(p x+q) \\sqrt{a x^{2}+b x+c} d x ; p \\neq 0, a \\neq 0<\/span> <\/b><\/p>\n

\u09a8\u09bf\u09df\u09ae (Rule): p x+q=m \\times \\frac{d}{d x}\\left(a x^{2}+b x+c\\right)+n, \\text { \u098f\u0996\u09be\u09a8\u09c7, } m=\\frac{p}{2 a} \\text {\u098f\u09ac\u0982 } n=q-\\frac{p b}{2 a}<\/span><\/b><\/p>\n

\u0985\u09b0\u09cd\u09a5\u09be\u09ce \u2018\u098f\u0995\u0998\u09be\u09a4 \u09b0\u09be\u09b6\u09bf\u09ae\u09be\u09b2\u09be\u09b0 \u09b8\u09cd\u09a5\u09b2\u09c7 \u09a6\u09cd\u09ac\u09bf\u0998\u09be\u09a4 \u09b0\u09be\u09b6\u09bf\u09ae\u09be\u09b2\u09be\u09b0 \u0985\u09a8\u09cd\u09a4\u09b0\u0995 \u09b8\u09b9\u0997\u2019 \u09b2\u09bf\u0996\u09c7 \u098f\u0995\u0998\u09be\u09a4 \u09b0\u09be\u09b6\u09bf\u09ae\u09be\u09b2\u09be\u09b0 \u09b8\u09ae\u09a4\u09be\u09ac\u09bf\u09a7\u09be\u09a8 \u0995\u09b0\u09a4\u09c7 \u09b9\u09df\u0964<\/span><\/p>\n

 <\/p>\n

\u0989\u09a6\u09be\u09b9\u09b0\u09a3-\u09ec: \\int \\frac{3 x+5}{\\sqrt{x^{2}+3 x}} d x<\/span><\/b>\u00a0\u09a8\u09bf\u09b0\u09cd\u09a3\u09df \u0995\u09b0\u0964<\/b><\/p>\n

Example \u2013 6: Determine \\int \\frac{3 x+5}{\\sqrt{x^{2}+3 x}} d x<\/span><\/b><\/p>\n

\u09b8\u09ae\u09be\u09a7\u09be\u09a8 (Solution):<\/b><\/p>\n\\int \\frac{3 x+5}{\\sqrt{x^{2}+3 x}} d x=\\int \\frac{\\frac{3}{2}(2 x+3)+5-\\frac{9}{2}}{\\sqrt{x^{2}+3 x}} d x<\/span>\n

 <\/p>\n

=\\int \\frac{3}{2}\\left\\{\\frac{2 x+3}{\\sqrt{x^{2}+3 x}}+\\frac{1}{2} \\frac{1}{\\sqrt{\\left(x+\\frac{3}{2}\\right)^{2}-\\left(\\frac{3}{2}\\right)^{2}}}\\right\\} d x<\/span><\/span><\/p>\n

=\\frac{3}{2} \\frac{\\left(x^{2}+3 x\\right)^{-\\frac{1}{2}+1}}{-\\frac{1}{2}+1}+\\ln \\left|\\sqrt{\\left(x+\\frac{3}{2}\\right)^{2}-\\left(\\frac{3}{2}\\right)^{2}}+\\left(x+\\frac{3}{2}\\right)\\right|+c<\/span><\/span><\/p>\n

=\\frac{3}{2} \\cdot 2\\left(x^{2}+3 x\\right)^{\\frac{1}{2}}+\\ln \\left|\\sqrt{x^{2}+3 x}+x+\\frac{3}{2}\\right|+c<\/span><\/span><\/p>\n

=3 \\sqrt{x^{2}+3 x}+\\ln \\left|\\sqrt{x^{2}+3 x}+x+\\frac{3}{2}\\right|+c<\/span><\/span>\u00a0 \u00a0(Ans)<\/b><\/p>\n

 <\/p>\n

\u09a7\u09b0\u09a8-\u09ea (Type \u2013 4):<\/b><\/p>\n\\int \\frac{p x^{2}+q x+r}{a x^{2}+b x+c} d x, \\int \\frac{p x^{2}+q x+r}{\\sqrt{a x^{2}+b x+c}} d x, \\int \\frac{\\alpha x+\\beta}{\\gamma x+\\delta} d x, \\int \\frac{a x+\\beta}{\\sqrt{\\gamma x+\\delta}} d x, \\int \\frac{a x+\\beta}{(\\gamma x+\\delta)^{n}} d x<\/span>\n

 <\/p>\n

\u09a8\u09bf\u09df\u09ae (Rule):<\/b> \u09b2\u09ac\u09c7\u09b0 \u09b0\u09be\u09b6\u09bf\u09ae\u09be\u09b2\u09be\u09b0 \u0998\u09be\u09a4 \u09b9\u09b0\u09c7\u09b0 \u09b0\u09be\u09b6\u09bf\u09ae\u09be\u09b2\u09be\u09b0 \u0998\u09be\u09a4\u09c7\u09b0 \u09b8\u09ae\u09be\u09a8 \u09ac\u09be \u09ac\u09c7\u09b6\u09bf \u09b9\u09b2\u09c7, \u09b2\u09ac\u09c7\u09b0 \u09b8\u09cd\u09a5\u09b2\u09c7 \u09b9\u09c1\u09ac\u09b9\u09c1 \u09b9\u09b0 \u09b2\u09bf\u0996\u09c7 \u09b8\u09ae\u09a4\u09be\u09ac\u09bf\u09a7\u09be\u09a8 \u0995\u09b0\u09be\u09b0 \u09aa\u09b0 \u09ad\u09be\u0997 \u0995\u09b0\u09c7 \u09a8\u09bf\u09a4\u09c7 \u09b9\u09df\u0964<\/span><\/p>\n

 <\/p>\n

\u0989\u09a6\u09be\u09b9\u09b0\u09a3-\u09ed: \\int \\frac{x d x}{\\sqrt{1-x}}<\/span><\/b>\u00a0\u09a8\u09bf\u09b0\u09cd\u09a3\u09df \u0995\u09b0\u0964<\/b><\/p>\n

Example 7 \u2013 Determine \\int \\frac{x d x}{\\sqrt{1-x}}<\/span><\/b><\/p>\n

\u09b8\u09ae\u09be\u09a7\u09be\u09a8 (Solution):<\/b><\/p>\n

\\int \\frac{x d x}{\\sqrt{1-x}}=\\int \\frac{-(1-x)+1}{\\sqrt{1-x}} d x<\/span> <\/span><\/p>\n=\\int \\frac{-(1-x)}{\\sqrt{1-x}} d x+\\int \\frac{1}{\\sqrt{1-x}} d x<\/span>\n=\\int-\\sqrt{1-x} d x+\\int \\frac{1}{\\sqrt{1-x}} d x<\/span>\n=+\\int(1-x)^{\\frac{1}{2}} d(1-x)-\\int \\frac{1}{\\sqrt{1-x}} d(1-x)<\/span>\n=\\frac{(1-x)^{\\frac{1}{2}+1}}{\\frac{1}{2}+1}-2 \\sqrt{1-x}+c<\/span>\n

 <\/p>\n

=\\frac{2}{3}(1-x)^{\\frac{3}{2}}-2 \\sqrt{1-x}+c<\/span>\u00a0 \u00a0 (Ans)<\/b><\/p>\n

 <\/p>\n

\u09a7\u09b0\u09a8-\u09eb: \\int \\frac{\\sqrt{a x+b}}{\\sqrt{c x+d}} d x<\/span><\/b><\/p>\n

Type \u2013 5: \\int \\frac{\\sqrt{a x+b}}{\\sqrt{c x+d}} d x<\/span><\/b><\/p>\n

\u09a8\u09bf\u09df\u09ae (Rule):<\/b> \u09b2\u09ac \u0993 \u09b9\u09b0\u0995\u09c7 \u09b2\u09ac \u09a6\u09cd\u09ac\u09be\u09b0\u09be \u0997\u09c1\u09a3 \u0995\u09b0\u09c7 \u09b2\u09ac\u0995\u09c7 <\/span>\u221a<\/span>\u00a0\u09ae\u09c1\u0995\u09cd\u09a4 \u0995\u09b0\u09a4\u09c7 \u09b9\u09ac\u09c7\u0964<\/span><\/p>\n

 <\/p>\n

\u0989\u09a6\u09be\u09b9\u09b0\u09a3-\u09ee: \\int \\sqrt{\\frac{a+x}{x}} d x<\/span> <\/b>\u09a8\u09bf\u09b0\u09cd\u09a3\u09df \u0995\u09b0\u0964<\/b><\/p>\n

Example \u2013 8: Determine \\int \\sqrt{\\frac{a+x}{x}} d x<\/span><\/b><\/p>\n

\u09b8\u09ae\u09be\u09a7\u09be\u09a8 (Solution):<\/b><\/p>\n\\int \\sqrt{\\frac{a+x}{x}} d x=\\int \\frac{(\\sqrt{a+x})^{2}}{\\sqrt{x(a+x)}} d x<\/span>\n

 <\/p>\n

=\\int \\frac{a+x}{\\sqrt{x^{2}+a x}} d x<\/span><\/span><\/p>\n

=\\int \\frac{\\frac{1}{2}(2 x+a)+\\frac{a}{2}}{\\sqrt{x^{2}+a x}} d x<\/span><\/span><\/p>\n

=\\frac{1}{2} \\int \\frac{2 x+a}{\\sqrt{x^{2}+a x}} d x+\\frac{a}{2} \\int \\frac{d x}{\\sqrt{\\left(x+\\frac{a}{2}\\right)^{2}-\\left(\\frac{a}{2}\\right)^{2}}}<\/span><\/span><\/p>\n

=\\frac{1}{2} \\cdot 2 \\sqrt{x^{2}+a x}+\\frac{a}{2} \\ln \\left|\\sqrt{\\left(x+\\frac{a}{2}\\right)^{2}-\\left(\\frac{a}{2}\\right)^{2}}+x+\\frac{a}{2}\\right|+c<\/span><\/span><\/p>\n

=\\sqrt{x^{2}+a x}+\\frac{a}{2} \\ln \\left|\\sqrt{x^{2}+a x}+x+\\frac{a}{2}\\right|+c<\/span>\u00a0 \u00a0<\/span>(Ans)<\/b><\/p>\n

 <\/p>\n

\u09a7\u09b0\u09a8-\u09ec: \\int \\frac{1}{g(x) \\sqrt{\\varphi(x)}} d x<\/span><\/b><\/p>\n

(a) <\/span>g(x)<\/span> \u0993 <\/span>\u03c6(x)<\/span> \u0989\u09ad\u09df\u09c7 \u098f\u0995\u0998\u09be\u09a4 \u09b9\u09b2\u09c7, \u03c6(<\/span>x) <\/span>= z^{2}<\/span> <\/span>\u00a0\u09a7\u09b0\u09a4\u09c7 \u09b9\u09df\u0964<\/span><\/p>\n

(b) <\/span>g(x)<\/span> \u098f\u0995\u0998\u09be\u09a4 \u0993 <\/span>\u03c6(x)<\/span> \u09a6\u09cd\u09ac\u09bf\u0998\u09be\u09a4 \u09b9\u09b2\u09c7, <\/span>g(<\/span>x) <\/span>= \\frac{1}{z}<\/span> <\/span>\u00a0\u09a7\u09b0\u09a4\u09c7 \u09b9\u09df\u0964<\/span><\/p>\n

(c) <\/span>g(x)<\/span> \u09a6\u09cd\u09ac\u09bf\u0998\u09be\u09a4 \u0993 <\/span>\u03c6(x)<\/span> \u098f\u0995\u0998\u09be\u09a4 \u09b9\u09b2\u09c7, \u03c6(x)<\/span>\u00a0<\/span>= z^{2}<\/span><\/span>\u00a0\u09a7\u09b0\u09a4\u09c7 \u09b9\u09df\u0964<\/span><\/p>\n

(d) <\/span>g(x)<\/span> \u0993 <\/span>\u03c6(x)<\/span> \u0989\u09ad\u09df\u09c7 \u09a6\u09cd\u09ac\u09bf\u0998\u09be\u09a4 \u09b9\u09b2\u09c7, <\/span>x =<\/span> \\frac{1}{z}<\/span> \u09a7\u09b0\u09a4\u09c7 \u09b9\u09df\u0964<\/span><\/p>\n

(e) \\int \\frac{x}{g(x) \\sqrt{\\varphi(x)}} d x<\/span><\/span>\u00a0\u098f\u09ac\u0982 <\/span>g(x)<\/span> \u0993 <\/span>\u03c6(x)<\/span> \u0989\u09ad\u09df\u09c7 \u09a6\u09cd\u09ac\u09bf\u0998\u09be\u09a4 \u09b9\u09b2\u09c7, <\/span>\u03c6(x) = z^{2}<\/span> <\/span>\u00a0\u09a7\u09b0\u09a4\u09c7 \u09b9\u09df\u0964<\/span><\/p>\n

 <\/p>\n

\u0989\u09a6\u09be\u09b9\u09b0\u09a3-\u09ef: \\int \\frac{d x}{(1-x) \\sqrt{1-x^{2}}}<\/span>\u00a0 <\/b>\u09a8\u09bf\u09b0\u09cd\u09a3\u09df \u0995\u09b0\u0964<\/b><\/p>\n

(Example \u2013 9: Determine \\int \\frac{d x}{(1-x) \\sqrt{1-x^{2}}}<\/span>)<\/b><\/p>\n

\u09b8\u09ae\u09be\u09a7\u09be\u09a8 (Solution):<\/b><\/p>\n

\u09a7\u09b0\u09bf, I=\\int \\frac{d x}{(1-x) \\sqrt{1-x^{2}}} \\text { \u098f\u09ac\u0982 } 1-x=\\frac{1}{z}, \\text { \u09a4\u09be\u09b9\u09b2\u09c7, } z=\\frac{1}{1-x} \\text { \u098f\u09ac\u0982 }-d x=-\\frac{1}{z^{2}} d z<\/span><\/span><\/p>\n

\\therefore I=\\int \\frac{d z}{z^{2} \\cdot \\frac{1}{z} \\sqrt{1-\\left(1-\\frac{1}{z}\\right)^{2}}}=\\int \\frac{d z}{z \\sqrt{1-1+2 \\frac{1}{z}-\\frac{1}{z^{2}}}}<\/span> <\/span><\/p>\n

=\\int \\frac{d z}{\\sqrt{2 z-1}}<\/span><\/span><\/p>\n

=\\frac{1}{2} \\int \\frac{d(2 z-1)}{\\sqrt{2 z-1}}<\/span><\/span><\/p>\n

=\\frac{1}{2} \\cdot 2 \\sqrt{2 z-1}+c<\/span><\/span><\/p>\n

\\therefore \\int \\frac{d x}{(1-x) \\sqrt{1-x^{2}}}=\\sqrt{2 \\cdot \\frac{1}{1-x}-1}+c<\/span> <\/span><\/p>\n=\\sqrt{\\frac{2-1+x}{1-x}}+c<\/span>\n

 <\/p>\n

=\\sqrt{\\frac{1+x}{1-x}}+c<\/span>\u00a0 \u00a0 \u00a0(Ans)<\/b><\/p>\n

 <\/p>\n

\u09a7\u09b0\u09a8-\u09ed (Type – 7):<\/b><\/p>\n

(a) \u09af\u09a6\u09bf \u0995\u09cb\u09a8\u09cb \u09af\u09cb\u0997\u099c \\int \\frac{a+b x^{l}}{p+q x^{m}} d x<\/span><\/span>\u00a0\u0986\u0995\u09be\u09b0\u09c7 \u09a5\u09be\u0995\u09c7, \u09af\u09c7\u0996\u09be\u09a8\u09c7 <\/span>l<\/span> \u0993 <\/span>m<\/span> \u0989\u09ad\u09df\u09c7 \u09ad\u0997\u09cd\u09a8\u09be\u0982\u09b6 \u098f\u09ac\u0982 \u09a4\u09be\u09a6\u09c7\u09b0 \u09b9\u09b0\u09c7\u09b0 \u09b2.\u09b8\u09be.\u0997\u09c1 <\/span>n<\/span> \u09b9\u09df, \u09a4\u09ac\u09c7 <\/span>x = z^{n}<\/span><\/span>\u00a0\u09a7\u09b0\u09a4\u09c7 \u09b9\u09df\u0964<\/span><\/p>\n

(b) \\int \\frac{d x}{x\\left(a+b x^{n}\\right)}<\/span><\/span> \u0986\u0995\u09be\u09b0\u09c7\u09b0 \u09af\u09cb\u0997\u099c\u09c7\u09b0 \u099c\u09a8\u09cd\u09af x^{n}=\\frac{1}{z}<\/span><\/span>\u00a0\u09a7\u09b0\u09a4\u09c7 \u09b9\u09df\u0964<\/span><\/p>\n

(c) \\int \\frac{d x}{x \\sqrt{a+b x^{n}}}<\/span> <\/span>\u0986\u0995\u09be\u09b0\u09c7\u09b0 \u09af\u09cb\u0997\u099c\u09c7\u09b0 \u099c\u09a8\u09cd\u09af x^{n}=\\frac{1}{z^{2}}<\/span><\/span>\u00a0\u09a7\u09b0\u09a4\u09c7 \u09b9\u09df\u0964<\/span><\/p>\n

(d) \\int \\frac{d x}{x^{m}(a+b x)^{n}}<\/span><\/span> \u0986\u0995\u09be\u09b0\u09c7\u09b0 \u09af\u09cb\u0997\u099c\u09c7\u09b0 \u099c\u09a8\u09cd\u09af a+b x=z x<\/span><\/span>\u00a0\u09a7\u09b0\u09a4\u09c7 \u09b9\u09df\u0964<\/span><\/p>\n

(e) \\int \\frac{d x}{(x-a)^{m}(x-b)^{n}}<\/span><\/span> \u0986\u0995\u09be\u09b0\u09c7\u09b0 \u09af\u09cb\u0997\u099c\u09c7\u09b0 \u099c\u09a8\u09cd\u09af z=\\frac{x-b}{x-a}<\/span><\/span>\u00a0\u09a7\u09b0\u09a4\u09c7 \u09b9\u09df\u0964<\/span><\/p>\n

 <\/p>\n

\u0989\u09a6\u09be\u09b9\u09b0\u09a3-\u09e7\u09e6: \\int \\frac{d x}{x^{1 \/ 2}-x^{1 \/ 4}}<\/span><\/b>\u00a0\u09a8\u09bf\u09b0\u09cd\u09a3\u09df \u0995\u09b0\u0964<\/b><\/p>\n

Example 10 \u2013 Determine \\int \\frac{d x}{x^{1 \/ 2}-x^{1 \/ 4}}<\/span><\/b><\/p>\n

\u09b8\u09ae\u09be\u09a7\u09be\u09a8 (Solution):<\/b><\/p>\n

\u098f\u0996\u09be\u09a8\u09c7, <\/span>2<\/span> \u0993 <\/span>4<\/span> \u098f\u09b0 \u09b2.\u09b8\u09be.\u0997\u09c1 <\/span>4<\/span>\u0964 \u09ae\u09a8\u09c7 \u0995\u09b0\u09bf, x=z^{4} \\quad \\therefore x^{1 \/ 4}=z \\text\u00a0 {\u098f\u09ac\u0982 }\u00a0 d x=4 z^{3} d z<\/span><\/span><\/p>\n

\u2234\\int \\frac{d x}{x^{1 \/ 2}-x^{1 \/ 4}}=\\int \\frac{4 z^{3} d z}{\\left(z^{4}\\right)^{1 \/ 2}-z}<\/span><\/span>\u00a0<\/span><\/p>\n

=\\int \\frac{4 z^{3} d z}{z^{2}-z}<\/span><\/span>\u00a0<\/span><\/p>\n

=4 \\int \\frac{z^{2}}{z-1} d z<\/span><\/span>\u00a0<\/span><\/p>\n

=4 \\int \\frac{z(z-1)+(z-1)+1}{z-1} d z<\/span><\/span>\u00a0<\/span><\/p>\n

=4 \\int\\left(z+1+\\frac{1}{z-1}\\right) d z<\/span><\/span><\/p>\n

=4\\left(\\frac{z^{2}}{2}+z+\\ln |z-1|\\right)+c<\/span><\/span><\/p>\n

=2 x^{1 \/ 2}+4 x^{1 \/ 4}+4 \\ln \\left|x^{1 \/ 4}-1\\right|+c<\/span><\/span>\u00a0 \u00a0 \u00a0(Ans)<\/b><\/p>\n

 <\/p>\n

\u09a7\u09b0\u09a8-\u09ee (Type \u2013 8):<\/b><\/p>\n

\\text { (a) } \\int \\frac{\\left(x^{2} \\pm 1\\right) d x}{a x^{4}+b x^{3}+c x^{2}+b x+a}, a \\neq 0<\/span><\/span><\/p>\n

\u09a8\u09bf\u09df\u09ae (Rule)<\/b>:<\/span> \\frac{d}{d x}\\left(x \\pm \\frac{1}{x}\\right)<\/span> \u09ac\u09b2\u09c7, \u09b2\u09ac \u0993 \u09b9\u09b0\u0995\u09c7 \\left(1 \\pm \\frac{1}{x^{2}}\\right) \\&\\left(x \\pm \\frac{1}{x}\\right)<\/span> <\/span>\u0986\u0995\u09be\u09b0\u09c7 \u09aa\u09cd\u09b0\u0995\u09be\u09b6 \u0995\u09b0\u09a4\u09c7 \u09b9\u09df\u0964<\/span><\/p>\n\\text { (b) } \\int \\frac{x^{2} d x}{a x^{4}+b x^{2}+c}, a \\neq 0<\/span>\n

 <\/p>\n

\u09a8\u09bf\u09df\u09ae (Rule)<\/b>: x^{2} \\text { \u098f\u09b0 \u09b8\u09cd\u09a5\u09b2\u09c7 } \\frac{1}{2 \\sqrt{a}}\\left\\{\\left(\\sqrt{a} x^{2}+\\sqrt{c}\\right)+\\left(\\sqrt{a} x^{2}-\\sqrt{c}\\right)\\right\\}<\/span><\/span> \u00a0\u09b2\u09bf\u0996\u09a4\u09c7 \u09b9\u09df\u0964<\/span><\/p>\n\\text { (c) } \\int \\frac{d x}{a x^{4}+b x^{2}+c}, a \\neq 0<\/span>\n

 <\/p>\n

\u09a8\u09bf\u09df\u09ae (Rule)<\/b>:<\/span> \u09b2\u09ac\u09c7 \\frac{1}{2 \\sqrt{c}}\\left\\{\\left(\\sqrt{a} x^{2}+\\sqrt{c}\\right)-\\left(\\sqrt{a} x^{2}- \\sqrt{c}\\right)\\right\\}<\/span><\/span>\u00a0\u09b2\u09bf\u0996\u09a4\u09c7 \u09b9\u09df\u0964<\/span><\/p>\n

 <\/p>\n

\u0989\u09a6\u09be\u09b9\u09b0\u09a3-\u09e7\u09e7:\\int \\frac{x^{-3 \/ 4}}{1+\\sqrt{x}} d x<\/span><\/b>\u00a0\u09a8\u09bf\u09b0\u09cd\u09a3\u09df \u0995\u09b0\u0964<\/b><\/p>\n

Example 11 \u2013 Determine \\int \\frac{x^{-3 \/ 4}}{1+\\sqrt{x}} d x<\/span><\/b><\/p>\n

\u09b8\u09ae\u09be\u09a7\u09be\u09a8 (Solution):<\/b><\/p>\n

\u09ae\u09a8\u09c7 \u0995\u09b0\u09bf, x=z^{4}<\/span><\/span> \u09a4\u09be\u09b9\u09b2\u09c7, d x=4 z^{3} d z<\/span><\/span><\/p>\n

\u2234\\int \\frac{x^{-3 \/ 4}}{1+\\sqrt{x}} d x=\\int \\frac{\\left(z^{4}\\right)^{-3 \/ 4}}{1+\\sqrt{z^{4}}} 4 z^{3} d z<\/span><\/span><\/p>\n

=\\int \\frac{z^{-3}}{1+z^{2}} 4 z^{3} d z<\/span><\/span><\/p>\n

=4 \\int \\frac{d z}{1+z^{2}}<\/span><\/span>\u00a0<\/span><\/p>\n

=4 \\tan ^{-1} z+c<\/span><\/span><\/p>\n

=4 \\tan ^{-1}\\left(x^{1 \/ 4}\\right)+c<\/span>\u00a0 \u00a0<\/span>\u00a0(Ans)<\/b><\/p>\n

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\u098f\u0987\u099a\u098f\u09b8\u09b8\u09bf \u0993 \u098f\u09a1\u09ae\u09bf\u09b6\u09a8 \u09aa\u09b0\u09c0\u0995\u09cd\u09b7\u09be\u09b0\u09cd\u09a5\u09c0\u09a6\u09c7\u09b0 \u099c\u09a8\u09cd\u09af \u0986\u09ae\u09be\u09a6\u09c7\u09b0 \u0995\u09cb\u09b0\u09cd\u09b8\u09b8\u09ae\u09c2\u09b9\u0983<\/strong><\/em><\/p>\n<\/div>\n<\/div>\n<\/div>\n