In the given equation as follows, use partial fractions to find the indefinite integral :-

see the equation as attached here

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

see the equation as attached here

iohanetc
2021-10-19
Answered

In the given equation as follows, use partial fractions to find the indefinite integral :-

see the equation as attached here

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

see the equation as attached here

You can still ask an expert for help

toroztatG

Answered 2021-10-20
Author has **98** answers

Step 1

We have to evaluate the following integral using partial fractions:

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

$=\int \left(\frac{{x}^{2}-2x+1+2x-1}{{x}^{2}-2x+1}\right)dx$

$=\int \left(\frac{{x}^{2}-2x+1}{{x}^{2}-2x+1}\right)dx+\int \left(\frac{2x}{{x}^{2}-2x+1}\right)dx-\int \left(\frac{1}{{x}^{2}-2x+1}\right)dx$

$=\int dx+\int \frac{2xdx}{{(x-1)}^{2}}-\int \frac{dx}{{(x-1)}^{2}}$

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

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

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

$=x+2\mathrm{ln}|x-1|-\frac{1}{x-1}+C$

We have to evaluate the following integral using partial fractions:

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