How to solve $\int \frac{\tan^{-1}x}{(1+x)^2}dx$?

Question

I know how to solve the following integral $$\int \frac{\tan^{-1}x}{(1+x^2)}dx$$ . We have to substitute $\tan^{-1}x$ as $t$ and we will be done.

After this one, I tried to find out $$\int \frac{\tan^{-1}x}{(1+x)^2}dx $$ but unfortunately I could not figure out how to solve it.

What I thought was, substitution $x=\tan^2 t$ but could not complete the track. Would you please help me on this regard?

Thanks in advance

Answer 1

Integrate by parts, $$\int \frac{\arctan x}{(1+x)^2}dx=\arctan x\int\frac{dx}{(1+x)^2}-\int\left(\frac{d(\arctan x)}{dx}\int\frac{dx}{(1+x)^2}\right)dx$$

$$=-\frac{\arctan x}{1+x}+\int\frac1{(1+x)(1+x^2)}dx$$

Now use Partial Fraction Decomposition, $$\frac1{(1+x)(1+x^2)}=\frac A{1+x}+\frac{Bx+C}{1+x^2}$$

Answer 2

$$\begin{align*} & \int \frac{\tan^{-1}(x)}{(1+x)^2} \, dx \\ &= -2 \int \frac{\tan^{-1}\left(\frac{1-y}{1+y}\right)}{\left(1+\frac{1-y}{1+y}\right)^2} \, \frac{dy}{(1+y)^2} \tag1 \\ &= -\frac12 \int \left(\frac\pi4 - \tan^{-1}(y)\right) \, dy \tag2 \\ &= - \frac{\pi}8 y + \frac12 \left(y \tan^{-1}(y) - \frac12 \ln\left(y^2+1\right)\right) + C \tag3 \\ &= - \frac{\pi}8 \frac{1-x}{1+x} + \frac12 \left(\frac{1-x}{1+x} \tan^{-1}\left(\frac{1-x}{1+x}\right) - \frac12 \ln\left(\frac{(1-x)^2}{(1+x)^2}+1\right)\right) + C \\ &= \frac12 \frac{x-1}{x+1} \tan^{-1}(x) - \frac14 \ln\left(\frac{x^2+1}{(x+1)^2}\right) + C \\ &= \frac12 \tan^{-1}(x) - \frac{\tan^{-1}(x)}{x+1} + \ln\sqrt[4]{1+\frac{2x}{x^2+1}} + C \end{align*}$$

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• $(1)$ : substitute $x=\dfrac{1-y}{1+y}$
• $(2)$ : simplify, assuming $y>-1$ so that $\tan^{-1}\left(\dfrac{1-y}{1+y}\right) = \dfrac\pi4-\tan^{-1}(y)$
• $(3)$ : integrate by parts