Question #03e6a

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Answer 1 · 2017-04-12T23:59:25

$=1/2 (tan^(-1) x+x/(1+x^2))$

$int_0^(tan^(-1) x)1/(1+tan^2 t) dt$

By Trigonometric Identities,

$=int_0^(tan^(-1) x)1/sec^2 t dt =int_0^(tan^(-1) x)cos^2 t dt =int_0^(tan^(-1)x)1/2(1+cos 2t)dt$

By integrating and $sin 2t=2sin t cos t$

$=1/2 [1+(sin 2t)/2]_0^(tan^(-1) x) =1/2[t+sin t cos t]_0^(tan^(-1) x)$

$=1/2(tan^(-1) x +sin(tan^(-1) x) cdot cos(tan^(-1) x))$

$=1/2(tan^(-1) x+ x/sqrt(1+x^2) cdot 1/sqrt(1+x^2))$

$=1/2(tan^(-1) x+x/(1+x^2))$

I hope that this was clear.