Gregory Jones

Answered

2022-01-03

Evaluate the following integrals.

$\int \frac{81}{{x}^{3}-9{x}^{2}}dx$

Answer & Explanation

Joseph Fair

Expert

2022-01-04Added 34 answers

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,

$\int \frac{81dx}{{x}^{3}-9{x}^{2}}=81\int \frac{dx}{{x}^{2}(x-9)}$

$=81\int [-\frac{1}{81x}-\frac{1}{9{x}^{2}}+\frac{1}{81(x-9)}]dx$

$=81[-\frac{1}{81}\mathrm{ln}\left|x\right|+\frac{1}{9x}+\frac{1}{81}\mathrm{ln}|x-9|]+c$

$(\because \int \frac{dx}{x-a}=\mathrm{ln}|x-a|+c,\int {x}^{n}dx=\frac{{x}^{n+1}}{n+1}+c$

$=-\mathrm{ln}\left|x\right|+\frac{9}{x}+\mathrm{ln}|x-9|+c)$

Step 2

hence, given integral is$\mathrm{ln}|x-9|-\mathrm{ln}\left|x\right|+\frac{9}{x}+c$ .

Given:

for evaluating given integral, we first simplify it then integrate is

so,

Step 2

hence, given integral is

redhotdevil13l3

Expert

2022-01-05Added 30 answers

Solution:

Vasquez

Expert

2022-01-07Added 457 answers

Let's represent it in the form:

We use the method of decomposition into the elementary elements. Let us expand the function into the simplest terms:

x:-9A+B=0

1: -9B=81

Solving it, we find:

A=-1; B=-9; C=1

We calculate the tabular integral: We

calculate the tabular integral: We

calculate the tabular integral:

Answer:

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