Find out the answer to "In the given equation as follows, use partial..."

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} - 2 x + 1} d x

toroztatG

Skilled2021-10-20Added 98 answers

Step 1
We have to evaluate the following integral using partial fractions:

\int \frac{x^{2}}{x^{2} - 2 x + 1} d x

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

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

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

= \int d x + \int [\frac{(x - 1) + (x - 1)}{{(x - 1)}^{2}}] d x + \int \frac{2 d x}{{(x - 1)}^{2}} - \int \frac{d x}{{(x - 1)}^{2}}

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

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

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

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