# Compute $\int_{0}^{1} \frac{\ln(x)}{(x+1)^2} \mathrm dx$

Compute $\int_{0}^{1} \frac{\ln(x)}{(x+1)^2} \mathrm dx$ any idea behind this ?

• did you try integration by parts? – user265328 Sep 10 '15 at 17:51
• Do it by parts. – user249332 Sep 10 '15 at 17:51

\begin{align} \int_0^1\frac{\ln x}{(1+x)^2}dx&=-\int_0^1\int_x^1\frac1y\frac1{(1+x)^2}dydx\\ &=-\int_0^1\int_0^y\frac1y\frac1{(1+x)^2}dxdy\\ &=-\int_0^1\frac1y\frac{y}{1+y}dy\\ &=-\int_0^1\frac{1}{1+y}dy\\ &=-\ln2 \end{align}

• What's the justification for the first equality? (Just wanting to learn). – YoTengoUnLCD Sep 10 '15 at 23:43
• @YoTengoUnLCD I used $\ln x =-\int_x^1\frac{dy}{y}$; does this make it clear? :-) – Math-fun Sep 11 '15 at 7:47

Hint. You may integrate by parts over $[a,1]$ ($0<a<1$): \begin{align} \int_a^{1} \frac{\ln(x)}{(x+1)^2} \mathrm dx&=\left.-\frac1{x+1}\ln x \right|_a^1+\int_a^1\frac1{x(x+1)}\:\mathrm dx \end{align} then use a partial fraction decomposition $$\frac1{x(x+1)}=\frac A{x}+\frac B{x+1}$$ to conclude with $a \to 0^+$.

Let $$\displaystyle I = \int_{0}^{1}\frac{\ln(x)}{(1+x)^2}dx = \lim_{a\rightarrow 0}\int_{a}^{1}\frac{\ln(x)}{(1+x)^2}dx$$

Now Using Integration By parts, We Get $$\displaystyle I = -\lim_{a\rightarrow 0}\left[\ln(x)\cdot \frac{1}{x+1}\right]_{a}^{1}+\lim_{a\rightarrow 0}\int_{a}^{1} \frac{1}{x(x+1)}dx$$

So we get $$\displaystyle I = -\lim_{a\rightarrow 0}\left[\ln(x)\cdot \frac{1}{x+1}\right]_{a}^{1}+\lim_{a\rightarrow 0}\int_{a}^{1}\left[\frac{1}{x}-\frac{1}{x+1}\right]$$

So Integral $$\displaystyle I = -\lim_{a\rightarrow 0}\left[\ln(x)\cdot \frac{1}{x+1}\right]_{a}^{1}+\lim_{a\rightarrow 0}\left[\ln|x|-\ln|x+1|\right]_{a}^{1}$$

Here we have Calculate for $a>0$

So $$\displaystyle I = \lim_{a\rightarrow 0^{+}}\frac{\ln(a)}{a+1}-\ln(2)+\lim_{a\rightarrow 0^{+}}\left[\ln(a+1)-\ln(a)\right]=\lim_{a\rightarrow 0^{+}}\frac{\ln(a)}{a+1}-\ln(2)+\lim_{a\rightarrow 0^{+}}\frac{\ln(a+1)}{\ln(a)}$$

So Using $\bf{L\; Hopital \; Rule}\;,$ We get $$\displaystyle \lim_{a\rightarrow 0^{+}}\frac{1}{a}-\ln(2)+\lim_{a\rightarrow 0^{+}}\frac{\ln(a)}{a+1}-\ln(2)+\lim_{a\rightarrow 0^{+}}\frac{a}{a+1}$$

So we get $$\displaystyle I = 0-\ln(2)+0 = -\ln(2)$$

• @mounirbensalem, funny comment. – user249332 Sep 10 '15 at 18:02
• i have deleted my comment ;) – mounir ben salem Sep 10 '15 at 18:25
• Your method is correct, and so is your answer; but there seem to be several mistakes in the last 3 lines of your solution. – user84413 Sep 12 '15 at 16:05

Alternative way. We know that $$\sum_{k\geq0}\left(-1\right)^{k}x^{k}=\frac{1}{1+x}$$ hence if we differentiate $$\sum_{k\geq1}\left(-1\right)^{k}kx^{k-1}=-\frac{1}{\left(1+x\right)^{2}}$$ so $$\int_{0}^{1}\frac{\log\left(x\right)}{\left(1+x\right)^{2}}=-\sum_{k\geq1}\left(-1\right)^{k}k\int_{0}^{1}x^{k-1}\log\left(x\right)dx$$ and integrating by parts $$\int_{0}^{1}x^{k-1}\log\left(x\right)dx=\left.\frac{x^{k}}{k}\log\left(x\right)\right|_{0}^{1}-\frac{1}{k}\int_{0}^{1}x^{k-1}dx=-\frac{1}{k^{2}}$$ hence $$\int_{0}^{1}\frac{\log\left(x\right)}{\left(1+x\right)^{2}}dx=\sum_{k\geq1}\frac{\left(-1\right)^{k}}{k}=-\log\left(2\right).$$

Let $\displaystyle t=\frac{1}{x}$ to get $\;\;\displaystyle-\int_1^{\infty}\frac{\ln t}{(t+1)^2}dt;\;\;$then let $u=\ln t, \; dv=\frac{1}{(t+1)^2}dt$ to get

$\displaystyle-\bigg[-\frac{\ln t}{t+1}+\int\left(\frac{1}{t}-\frac{1}{t+1}\right)dt\bigg]_1^{\infty}=\lim_{b\to\infty}\bigg[\frac{\ln t}{t+1}-\ln\left(\frac{t}{t+1}\right)\bigg]_1^b$

$\displaystyle=\lim_{b\to\infty}\left(\frac{\ln b}{b+1}-\ln\frac{b}{b+1}+\ln\frac{1}{2}\right)=\ln\frac{1}{2}=-\ln2$