lugreget9

2021-12-28

Evaluate the integral.
$\int \frac{x+1}{{x}^{2}+2x}dx$

aquariump9

Step 1
Given: $I=\int \frac{x+1}{{x}^{2}+2x}dx$
for evaluating given integral, we substitute
${x}^{2}+2x=t$...(1)
now, differentiating equation (1) with respect to x
$\frac{d}{dx}\left({x}^{2}+2x\right)=\frac{d}{dx}\left(t\right)$

$2x+2\left(1\right)=\frac{dt}{dx}$
$2x+2=\frac{dt}{dx}$
$2\left(x+1\right)=\frac{dt}{dx}$
$\left(x+1\right)dx=\frac{dt}{2}$
Step 2
now, replacing (x+1)dx with $\frac{dt}{2},\left({x}^{2}+2x\right)$ with t in given integral
so,

$=\frac{1}{2}\mathrm{ln}|t|+c$...(2)
now replacing t with $\left({x}^{2}+2x\right)$ in equation (2)
so,
$\int \frac{x+1}{{x}^{2}+2x}dx=\frac{1}{2}\mathrm{ln}|{x}^{2}+2x|+c$
hence, given integral is equal to $\frac{1}{2}\mathrm{ln}|{x}^{2}+2x|+c$.

porschomcl

$\int \frac{x+1}{{x}^{2}+2x}dx$
Lets

Vasquez

$\int \frac{x+1}{{x}^{2}+2x}dx$
Transform
$\int \frac{1}{2t}dt$
Use properties
$\frac{1}{2}×\int \frac{1}{t}dt$
$\frac{1}{2}\ast \mathrm{ln}\left(|t|\right)$
Substitute back
$\frac{1}{2}\ast \mathrm{ln}\left(|{x}^{2}+2x|\right)$
$\frac{1}{2}\ast \mathrm{ln}\left(|{x}^{2}+2x|\right)+C$