Question

interagration of tan^-1 2x / 1+x^2 x= 1 and 0​

Answer 1

Let  x=tanθ , so that  θ=tan−1x ,  dx=sec2θdθ  

Then the given integral is equivalent to

∫tan−12x1−x2dx=∫tan−12tanθ1−tan2θ⋅sec2θdθ  

=∫tan−1tan2θ⋅sec2θdθ  

=2×∫θsec2θdθ  

(integrate by parts)

=2θ⋅∫sec2θdθ−2⋅∫1⋅(∫sec2θdθ)dθ  

=2θtanθ−2⋅∫tanθdθ  

=2θtanθ−2logesecθ+c  

=2xtan−1x−2loge1+x2−−−−−√+c  

=2xtan−1x−loge(1+x2)+c  

An Easier Method

With the substitution  x=tanθ , the integrand reduces to  2tan−1x . Therefore the given integral

=∫2tan−1xdx=2⋅∫tan−1x⋅1dx  

(integrate by parts)

=2tan−1x⋅x−2⋅∫11+x2⋅xdx  

=2xtan−1x−∫2x1+x2dx  

=2xtan−1x−loge(1+x2)+c

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