# In the given equation as follows, use partial fractions to find the indefinite i

In the given equation as follows, use partial fractions to find the indefinite integral :-
see the equation as attached here
$\int \frac{{x}^{2}}{{x}^{2}-2x+1}dx$
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

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