# Arctan integral

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Prove the following integral

$$\displaystyle \int^1_0 \frac{\mathrm{arctan}(x)}{1+x}\,dx = \frac{\pi}{8} \log(2)$$​

This is not too challenging and could be solved by elementary functions .

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#### Pranav

##### Well-known member
Prove the following integral

$$\displaystyle \int^1_0 \frac{\mathrm{arctan}(x)}{1+x}\,dx = \frac{\pi}{8} \log(2)$$​

This is not too challenging and could be solved by elementary functions .

Use the substitution $x=\tan t \Rightarrow dx=\sec^2t\,dt$, the integral changes to:

$$\int_0^{\pi/4} \frac{t\sec^2t}{1+\tan t}dt$$

From integration by parts and since $\displaystyle \int \frac{\sec^2t}{1+\tan t}dt=\ln(1+\tan t)+C$, we get

$$\displaystyle \bigg(t\ln(1+\tan t)\bigg|_0^{\pi/4}-\int_0^{\pi/4}\ln(1+\tan t)\,dt \,\,\, (*)$$

Let

$$I=\int_0^{\pi/4} \ln(1+\tan t)\,dt$$
The above is equivalent to
$$I=\int_0^{\pi/4} \ln\left(1+\tan\left(\frac{\pi}{4}-t\right)\right)\,dt$$
Adding both the expressions for I and simplifying, we get
$$2I=\int_0^{\pi/4}\ln2\,dt \Rightarrow I=\frac{\pi}{8}\ln2$$

Substituting in (*), we get the final answer $\displaystyle \frac{\pi}{8}\ln2$

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