Gregory Jones

2022-01-03

Evaluate the following integrals.
$\int \frac{81}{{x}^{3}-9{x}^{2}}dx$

### Answer & Explanation

Joseph Fair

Step 1
Given: $I=\int \frac{81}{{x}^{3}-9{x}^{2}}dx$
for evaluating given integral, we first simplify it then integrate is
so,
$\int \frac{81dx}{{x}^{3}-9{x}^{2}}=81\int \frac{dx}{{x}^{2}\left(x-9\right)}$
$=81\int \left[-\frac{1}{81x}-\frac{1}{9{x}^{2}}+\frac{1}{81\left(x-9\right)}\right]dx$
$=81\left[-\frac{1}{81}\mathrm{ln}|x|+\frac{1}{9x}+\frac{1}{81}\mathrm{ln}|x-9|\right]+c$
$\left(\because \int \frac{dx}{x-a}=\mathrm{ln}|x-a|+c,\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+c$
$=-\mathrm{ln}|x|+\frac{9}{x}+\mathrm{ln}|x-9|+c\right)$
Step 2
hence, given integral is $\mathrm{ln}|x-9|-\mathrm{ln}|x|+\frac{9}{x}+c$.

redhotdevil13l3

$\int \frac{81}{{x}^{3}-9{x}^{2}}dx$
$81\cdot \int \frac{1}{{x}^{3}-9{x}^{2}}dx$
$81\cdot \int -\frac{1}{81x}-\frac{1}{9{x}^{2}}+\frac{1}{81\left(x-9\right)}dx$
$81\left(-\int \frac{1}{81x}dx-\int \frac{1}{9{x}^{2}}+\frac{1}{81\left(x-9\right)}\right)dx$
$81\left(-\frac{1}{81}\cdot \mathrm{ln}\left(|x|\right)+\frac{1}{9x}+\frac{1}{81}\cdot \mathrm{ln}\left(|x-9|\right)\right)$
$-\mathrm{ln}\left(|x|\right)+\frac{9}{x}+\mathrm{ln}\left(|x-9|\right)$
Solution:
$-\mathrm{ln}\left(|x|\right)+\frac{9}{x}+\mathrm{ln}\left(|x-9|\right)+C$

Vasquez

$\int \frac{81}{{x}^{3}-9\ast {x}^{2}}dx$
Let's represent it in the form:
$\frac{81}{{x}^{2}\ast \left(x-9\right)}=\frac{81}{{x}^{2}\ast \left(x-9\right)}$
We use the method of decomposition into the elementary elements. Let us expand the function into the simplest terms:
$\frac{81}{{x}^{2}\left(x-9\right)}=\frac{A}{x}+\frac{B}{{x}^{2}}+\frac{C}{x-9}=\frac{Ax\left(x-9\right)+B\left(x-9\right)+C{x}^{2}}{{x}^{2}\left(x-9\right)}$
$81=Ax\left(x-9\right)+B\left(x-9\right)+C{x}^{2}$
${x}^{2}:A+C=0$
x:-9A+B=0
1: -9B=81
Solving it, we find:
A=-1; B=-9; C=1
$\frac{81}{{x}^{2}\left(x-9\right)}=\frac{-1}{x}+\frac{-9}{{x}^{2}}+\frac{1}{x-9}$
We calculate the tabular integral: We
$\int \frac{1}{x-9}dx=\mathrm{ln}\left(x-9\right)$
calculate the tabular integral: We
$-\int \frac{1}{x}dx=-\mathrm{ln}\left(x\right)$
calculate the tabular integral:
$\int -\frac{9}{{x}^{2}}dx=\frac{9}{x}$
$\mathrm{ln}\left(x-9\right)-\mathrm{ln}\left(x\right)+\frac{9}{x}+C$