\(\int\frac{(x)\left(e^{2x}\right)}{(1+2 x)^{2}}\) (where C is a constant of integration.)
\(\int\frac{(x)\left(e^{2x}\right)}{(1+2 x)^{2}}\)
(where C is a constant of integration.)
$\frac{e^{2 x}}{1+2 x}+C$
$\frac{e^{2 x}}{4(1+2 x)}+C$
$\frac{4 e^{2 x}}{1+2 x}+C$
$\frac{e^{2 x}}{2(1+2 x)}+C$
Solution
$\begin{aligned} & \frac{x}{(1+2 x)^2}=\frac{A}{(1+2 x)}+\frac{B}{(1+2 x)^2} \\ & \Rightarrow x=A(1+2 x)+B\end{aligned}$
For $x=-1 / 2$, equation: $-1 / 2=B$ i.e. $B=-1 / 2$
For $x=0$, equation: $0=A-1 / 2$ i.e. $A=1 / 2$
$\begin{aligned} & \therefore \frac{x}{(1+2 x)^2} \\ & =\frac{1}{2(1+2 x)}-\frac{1}{2(1+2 x)^2}\end{aligned}$
The given equation becomes
$\begin{aligned} & \int \mathrm{e}^{2 \mathrm{x}}\left[\frac{1}{2(1+2 \mathrm{x})}-\frac{1}{2(1+2 \mathrm{x})^2}\right] \mathrm{dx} \\ & =\int \mathrm{e}^{2 \mathrm{x}} \times \frac{1}{2(1+2 \mathrm{x})} \mathrm{dx}-\int \mathrm{e}^{2 \mathrm{x}} \times \frac{1}{2(1+2 \mathrm{x})^2} \mathrm{dx}\end{aligned}$
Tip - If $f_1(x)$ and $f_2(x)$ are two functions, then an integral of the form $f$ $\mathrm{f}_1(\mathrm{x}) \mathrm{f}_2(\mathrm{x}) \mathrm{dx}$ can be INTEGRATED BY PARTS as
$f_1(x) \int f_2(x) d x-\int\left\{\frac{d}{d x} f_1(x) \int f_2(x) d x\right\} d x$
where $f_1(x)$ and $f_2(x)$ are the first and second functions respectively.
Taking $f_1(x)=1 /(1+2 x)$ and $f_2(x)=e^{2 x}$ in the second integral and keeping the first integral intact,
$\begin{aligned}
& \begin{array}{l}
\int \mathrm{e}^{2 \mathrm{x}} \times \frac{1}{2(1+2 \mathrm{x})} \mathrm{dx} \\
-\int \mathrm{e}^{2 \mathrm{x}} \times \frac{1}{2(1+2 \mathrm{x})^2} \mathrm{dx}
\end{array} \\
& =\frac{1}{2}\left[\frac{1}{1+2 \mathrm{x}} \int \mathrm{e}^{2 \mathrm{x}} \mathrm{dx}-\int\left[\frac{\mathrm{d}}{\mathrm{dx}\left(\frac{1}{1+2 \mathrm{x}}\right)}\right.\right. \\
& \left.\left.\iint \mathrm{e}^{2 \mathrm{x}} \mathrm{dx}\right] \mathrm{dx}-\int \frac{\mathrm{e}^{2 \mathrm{x}}}{(1+2 \mathrm{x})^2} \mathrm{dx}\right] \\
& =\frac{1}{2}\left[\frac{\mathrm{e}^{2 \mathrm{x}}}{2(2 \mathrm{x}+1)}+\int \frac{\mathrm{e}^{2 \mathrm{x}}}{(2 \mathrm{x}+1)^2} \mathrm{dx}\right. \\
& \left.\quad-\int \frac{\mathrm{e}^{2 \mathrm{x}}}{(2 \mathrm{x}+1)^2} \mathrm{dx}\right] \\
& =\frac{\mathrm{e}^{2 \mathrm{x}}}{4(2 \mathrm{x}+1)}+\mathrm{c},
\end{aligned}$
$\text { where } \mathrm{c} \text { is the integrating constant. }$