TITLE: How do I find the following definite integral?
QUESTION [7 upvotes]: Find the value of the following definite integral:
$$\int_0^{1}\frac{\arctan(t)}{1+t}dt$$
I tried using integration by parts, but it gives another complicated integral, so possibly the antiderivative does not exist. What other methods can I apply?

REPLY [12 votes]: Let us write $$I=\int_0^1\dfrac{\arctan t}{1+t}dt$$
The change of variables  $t=\dfrac{1-x}{1+x}$ shows that
$$\eqalign{I&=\int_0^1\arctan\left(\frac{1-x}{1+x}\right)\frac{dx}{1+x}\\
&=\int_0^1\left(\frac{\pi}{4}-\arctan x\right)\frac{dx}{1+x}\\
&=\frac{\pi}{4}\int_0^1\frac{dx}{1+x}-\int_0^1\frac{\arctan x}{1+x}dx\\
&=\frac{\pi}{4}\ln 2-I
}$$
Thus $I=\dfrac{\pi}{8}\ln 2$.$\qquad \square$