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प्रश्न
Evaluate : `int tan^-1x dx`
उत्तर
Let I = `int tan^-1 x dx`
= `int tan^-1x . 1dx`
Integrating by parts, we get
I = `tan^-1 x int 1 dx - int [d/dx ( tan^-1 x) . int 1.dx] dx`
= `tan^-1 x . x - int 1/[ 1 + x^2 ]. x dx`
= `x tan^-1 x - 1/2 int 2x/[ 1 + x^2 ] dx`
= `x tan^-1 x - 1/2 log |( 1 + x^2)| + c`
`( ∵ int f'(x)/f(x) dx = log | f(x) | + c)`
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