MHB Prove: Integral of arctan(x) = $\frac{\pi}{8}\log(2)$

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The integral of arctan(x) over the interval from 0 to 1, specifically $$\int^1_0 \frac{\mathrm{arctan}(x)}{1+x}\,dx$$, is proposed to equal $$\frac{\pi}{8} \log(2)$$. Participants in the discussion agree that this integral can be approached using elementary functions, suggesting that the proof is accessible. Various methods for proving this equality are explored, emphasizing the integral's relationship with logarithmic and trigonometric functions. The consensus is that a straightforward solution exists without the need for advanced techniques. The discussion ultimately aims to validate the integral's value through clear mathematical reasoning.
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Prove the following integral

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

This is not too challenging and could be solved by elementary functions .
 
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ZaidAlyafey said:
Prove the following integral

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

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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