iohanetc

## Answered question

2021-10-19

In the given equation as follows, use partial fractions to find the indefinite integral :-
see the equation as attached here
$\int \frac{{x}^{2}}{{x}^{2}-2x+1}dx$

### Answer & Explanation

toroztatG

Skilled2021-10-20Added 98 answers

Step 1
We have to evaluate the following integral using partial fractions:
$\int \frac{{x}^{2}}{{x}^{2}-2x+1}dx$
$=\int \left(\frac{{x}^{2}-2x+1+2x-1}{{x}^{2}-2x+1}\right)dx$
$=\int \left(\frac{{x}^{2}-2x+1}{{x}^{2}-2x+1}\right)dx+\int \left(\frac{2x}{{x}^{2}-2x+1}\right)dx-\int \left(\frac{1}{{x}^{2}-2x+1}\right)dx$
$=\int dx+\int \frac{2xdx}{{\left(x-1\right)}^{2}}-\int \frac{dx}{{\left(x-1\right)}^{2}}$
$=\int dx+\int \left[\frac{\left(x-1\right)+\left(x-1\right)}{{\left(x-1\right)}^{2}}\right]dx+\int \frac{2dx}{{\left(x-1\right)}^{2}}-\int \frac{dx}{{\left(x-1\right)}^{2}}$
$=\int dx+\int \frac{dx}{x-1}+\int \frac{dx}{x-1}+\int \frac{dx}{{\left(x-1\right)}^{2}}$
$=\int dx+2\int \frac{dx}{x-1}+\int \frac{dx}{{\left(x-1\right)}^{2}}$
$=x+2\mathrm{ln}|x-1|-\frac{1}{x-1}+C$

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?