0
$\begingroup$

This question already has an answer here:

Find the value of following function's integral with respect to x from 0 to 1

$$F(x)=\frac{\ln(1+x)}{1+x^2}$$

$\endgroup$

marked as duplicate by Ron Gordon calculus Feb 15 '17 at 20:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

3
$\begingroup$

The hint: Let $x=\tan \theta$ and the rest is smooth: $$\int\limits_0^1\frac{\ln(1+x)}{1+x^2}dx-\frac{\pi}{8}\ln2=\int\limits_{0}^{\frac{\pi}{4}}\ln(1+\tan\theta)d\theta-\frac{\pi}{8}\ln2=$$ $$=\int\limits_{0}^{\frac{\pi}{4}}\ln\frac{\sqrt2\sin\left(\frac{\pi}{4}+\theta\right)}{\cos\theta}d\theta-\frac{\pi}{8}\ln2=$$ $$=\int\limits_{0}^{\frac{\pi}{4}}\ln\sin\left(\frac{\pi}{4}+\theta\right)d\theta-\int\limits_{0}^{\frac{\pi}{4}}\ln\cos\theta d\theta=$$ $$=\int\limits_{0}^{\frac{\pi}{4}}\ln\sin\left(\frac{\pi}{4}+\theta\right)d\theta-\int\limits_{\frac{\pi}{4}}^{0}\ln\sin\left(\frac{\pi}{4}+\frac{\pi}{4}-\theta\right) d\left(\frac{\pi}{4}-\theta\right)=0.$$ Id est, $$\int\limits_0^1\frac{\ln(1+x)}{1+x^2}dx=\frac{\pi}{8}\ln2$$

$\endgroup$
2
$\begingroup$

Let us consider the double integral

$$I=\int_{0}^{1} \int_{0}^{1} \frac{x}{(1+xy)(1+x^2)} \ dy \ dx.$$ Do the inner integral to get $$I= \int_{0}^{1} \frac{\ln(1+x)}{x^2+1} \ dx.$$

Now reverse the order of integration, justified by Fubini's Theorem.

$$I=\int_{0}^{1} \int_{0}^{1} \frac{x}{(1+xy)(1+x^2)} \ dx \ dy.$$ Now we use partial fractions on the integrand

$$\frac{x}{(1+xy)(1+x^2)}= \frac{x}{(x^2+1)(y^2+1)}+\frac{y}{(x^2+1)(y^2+1)} - \frac{y}{(1+xy)(y^2+1)},$$ and evaluate the double integral, which simplifies to

$$I= \int_{0}^{1} \frac{\ln(2)}{2(y^2+1)} + \frac{\pi y}{4(y^2+1)} - \frac{\ln(1+y)}{y^2+1} \ dy = \frac{\pi \ln(2)}{4}- I.$$

which gives us $$I=\frac{\pi \ln(2)}{8}.$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.