# Evaluate the following: $\int\limits_0^1 x\;log(1+2x)dx$

Toolbox:
• $\int u dv=uv-\int v du$
• If a rational expression $\large\frac{p(x)}{q(x)}$ where the degree of $p(x)$ is greater than $q(x)$,then it is said to be improper.
Step 1:
Let $I=\int_0^1 x\log(1+2x) dx.$
Clearly the integral is of the form $\int u dv.$
Hence this can be solved by the method of integration by parts
$\int u dv=uv-\int v du.$
Let $u=\log(1+2x)$.On differentiating with respect to x
$du=\large{\frac{1}{1+2x}}\normalsize .2dx.$
Let $dv=x dx$,hence on integrating we get
$v=\large\frac{x^2}{2}$
Now substituting for $u,v,du$ and $dv$ we get,
$\int_0^1 x\log(1+2x)dx=\bigg(\frac{x^2}{2}.\log(1+2x)\bigg)_0^1-\int_0^1\large\frac{x^2}{2}.\frac{2dx}{1+2x}$
$\qquad\qquad=\bigg(\frac{x^2}{2}.\log(1+2x)\bigg)_0^1-\int_0^1\large\frac{x^2dx}{1+2x}$
Consider $\int_0^1\large\frac{x^2dx}{1+2x}$
The expression $\frac{x^2}{1+2x}$ is improper.To make this a proper rational expression let us divide
Hence $\large\frac{x^2}{1+2x}=\big(\frac{x^2}{4}-\frac{1}{4}\big)+\frac{1/4}{1+2x}$.
$\qquad\qquad\quad=\large \frac{x^2}{4}-\frac{1}{4}+\frac{1}{4(1+2x)}$
Step 2:
Hence $\int_0^1\large\frac{x^2}{1+2x}=\int_0^1\large\frac{x^2}{4}-\frac{1}{4}\int_0^1 dx+\frac{1}{4}\int_0^1\large\frac{dx}{1+2x}$.
On integrating we get,
$\Rightarrow \large\frac{x^3}{12}-\large\frac{1}{4}\normalsize x+\large\frac{1}{4}\normalsize \log(1+2x).$
$I=\begin{bmatrix}\large\frac{x^2}{2}\normalsize \log(1+2x)\end{bmatrix}_0^1-\begin{bmatrix}\large\frac{x^3}{12}-\frac{x}{4}+\frac{1}{8}\log (1+2x).\frac{1}{2}\end{bmatrix}_0^1$
On applying limits,
$\;\;=\begin{bmatrix}\large\frac{1}{2}\normalsize \log(1+2\times 1)\end{bmatrix}-\begin{bmatrix}\large\frac{1}{12}-\frac{1}{4}+\frac{1}{8}\log (1+2\times 1)-0-0+\frac{1}{8}(1+0)\end{bmatrix}$
$\;\;=(\large\frac{1}{2}\normalsize \log 3)-(\large\frac{1}{8}\normalsize \log 3)$
$\;\;=\large\frac{3}{8}\normalsize \log 3$