How would you solve this integral?

**Christopher13** · July 2nd 2012, 09:04 AM

I have been trying to solve this integral for a while but nothing works. I tried using WolframAlpha.com but the answer is a very nasty one, and my Calculus book gives a nice integral for this, using x= (1-u)/(1+u), but why would they do that? The answer is (pi/8)ln(2). Thank you in advance!.

∫ ln(x+1)/(x^2+1) from 0 to 1

**thevinh** · July 4th 2012, 12:42 AM

Originally Posted by Christopher13

I have been trying to solve this integral for a while but nothing works. I tried using WolframAlpha.com but the answer is a very nasty one, and my Calculus book gives a nice integral for this, using x= (1-u)/(1+u), but why would they do that? The answer is (pi/8)ln(2). Thank you in advance!.

∫ ln(x+1)/(x^2+1) from 0 to 1

$\text{From the substitution we can find out some of the quantities that involved in the integral.}$

$x=\frac{1-u}{1+u}\, \Rightarrow\, u=\frac{1-x}{1+x}\quad \quad \text{(Solve for \emph{u})}$

$x=0\, \Rightarrow\, u=1\mspace{7mu} \text{and}\mspace{7mu} x=1\, \Rightarrow\, u=0$

$x=\frac{1-u}{1+u}\, \Rightarrow\, dx=-\frac{2}{(1+u)^2}\,du}$

$x+1=\frac{1-u}{1+u}+1\, \Leftrightarrow\, x+1=\frac{2}{1+u}$

$x^2+1=\left(\frac{1-u}{1+u}\right)^2+1\, \Leftrightarrow\, x^2+1=\frac{2(1+u^2)}{(1+u)^2}\, \Leftrightarrow\, \frac{1}{x^2+1}=\frac{(1+u)^2}{2(1+u^2)}$

$\begin{align*} \int_0^1{\frac{\ln{(x+1)}}{x^2+1}}\,dx &= \int_1^0{\frac{(1+u)^2}{2(1+u^2)} \cdot \ln{\left(\frac{2}{1+u}\right)} \cdot \left(-\frac{2}{(1+u)^2}\,du \right)}\\ &=-\int_1^0{\frac{\ln{(2)}-\ln{(u+1)}}{u^2+1}\,du}\\ &= \int_0^1{\frac{\ln{(2)}-\ln{(u+1)}}{u^2+1}\,du}\\ &=\ln{(2)}\int_0^1{\frac{du}{u^2+1}} - \int_0^1{\frac{\ln{(u+1)}}{u^2+1}\,du} \end{align*}$

$\text{Since the variables of integration are just dummy variables, we can say that:}\mspace{7mu}$ $\displaystyle \int_0^1{\frac{\ln{(x+1)}}{x^2+1}}}\,dx}=\int_0^1{ \frac{\ln{(u+1)}}{u^2+1}\,du}$

$\text{Now collect liked terms to the left.}$

$\begin{align*} 2\int_0^1{\frac{\ln{(x+1)}}{x^2+1}\,dx}&=\ln{(2)} \int_0^1{\frac{du}{u^2+1}}\\ &=\frac{\ln{(2)}}{2} \cdot \left. \arctan{u}\right|_0^1\\ &=\frac{\ln{(2)}}{2} \cdot \frac{\pi}{4}\\ &=\frac{\pi}{8}\ln{(2)} \end{align*}$

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