Evaluate the following integrals. ∫81x3−9x2dx

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

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

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