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AAKASH SERIES-INDEFINITE INTEGRALS -EXERCISE -II
- int Sin^(-1)((2x)/(1+x^(2)))dx=f(x)-log (1+x^(2))+c rArr f(x)=
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- int x Tan^(-1) x dx=
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- Evaluate int tan^(-1) sqrt((1-x)/(1+x)) dx, on (-1,1 )
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- int (sin^(-1) x -cos^(-1)x)/(sin^(-1) x + cos^(-1)x) dx =
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- int (x tan^(-1)x)/(sqrt(1 + x^(2))) d x=
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- int (sin^(-1) sqrt(x))/(sqrt(1 -x)) dx =
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- int e^(x)((1+sin x)/(1+cos x))dx=
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- int e^(x)((2+sin 2x)/(1+cos 2x)dx=
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- int e^(2x) (2sinx + cosx) dx =
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- int e^(x)(x+sqrt(1+x^(2)))(1+(1)/(sqrt(1+x^(2))))dx=
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- int (x.e^(x))/((2 + x)^(3)) dx =
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- int (e^(x)(x^(2)+1))/((1+x)^(2))dx=
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- int ((2-sin 2x)/(1-cos 2x))e^(x)dx=
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- int (3-x^(2))/(1-2x+x^(2))e^(x)dx=e^(x)f(x)+c rArr f(x)=
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- int e^(x).(x^(3) + x + 1)/((1 + x^(2))^(3//2)) dx =
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- int [ sin(log x)+ cos (log x ) ] dx =
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- int [ (1)/(log x) - (1)/((log x)^(2)) ] dx =
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- int [ (log x - 1)/(1 + (log x)^(2)) ]^(2) dx =
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- int e^(x) [ "In " x + (1)/(x^(2)) ] dx =
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- int [ log(log x) + (log x )^(-2) ] dx =
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