How do you integrate int (x^2-2x+1)/(x-2)^3 using partial fractions?

Jan 18, 2017

$\ln | x - 2 | - \frac{2}{x - 2} - \frac{1}{2 {\left(x - 2\right)}^{2}} + C .$

Explanation:

We can solve the Problem without decomposing the

Integrand into Partial Fractions as shown below :

$I = \int \frac{{x}^{2} - 2 x + 1}{x - 2} ^ 3 \mathrm{dx} = \int \frac{x \left(x - 2\right) + 1}{x - 2} ^ 3 \mathrm{dx}$

$= \int \left\{\frac{x \left(x - 2\right)}{x - 2} ^ 3 + \frac{1}{x - 2} ^ 3\right\} \mathrm{dx}$

$\int \left\{\frac{x}{x - 2} ^ 2 + \frac{1}{x - 2} ^ 3\right\} \mathrm{dx}$

$= \int \left[\frac{\left(x - 2\right) + 2}{x - 2} ^ 2 + \frac{1}{x - 2} ^ 3\right] \mathrm{dx}$

$= \int \left\{\frac{x - 2}{x - 2} ^ 2 + \frac{2}{x - 2} ^ 2 + \frac{1}{x - 2} ^ 3\right\} \mathrm{dx}$

$= \int \left\{\frac{1}{x - 2} + \frac{2}{x - 2} ^ 2 + \frac{1}{x - 2} ^ 3\right\} \mathrm{dx}$

Now, we use the following Result :

$\int f \left(x\right) \mathrm{dx} = F \left(x\right) + c \Rightarrow \int f \left(a x + b\right) = \frac{1}{a} F \left(a x + b\right) , a \ne 0$.

[ It can be easily proved by subst.ing, for $a x + b$.]

$\therefore I = \ln | x - 2 | - \frac{2}{x - 2} - \frac{1}{2 {\left(x - 2\right)}^{2}} + C .$

Enjoy Maths. , and, spread the Joy!

Jan 18, 2017

$\ln | x - 2 | - \frac{2}{x - 2} - \frac{1}{2 {\left(x - 2\right)}^{2}} + C .$

Explanation:

Noticing that ${\left(x - 2\right)}^{3}$ is a perfect cube, there is no need for partial fractions. Substitute $u = x - 2$ and $\frac{\mathrm{du}}{\mathrm{dx}} = 1$ and the integral becomes:
$\int \frac{{\left(u + 2\right)}^{2} - 2 \left(u + 2\right) + 1}{u} ^ 3 \times 1 \mathrm{du}$
$= \int \frac{{u}^{2} + 2 u + 1}{u} ^ 3 \mathrm{du}$
$= \int \frac{1}{u} + \frac{2}{u} ^ 2 + \frac{1}{u} ^ 3 \mathrm{du}$
$= \ln | x - 2 | - \frac{2}{x - 2} - \frac{1}{2 {\left(x - 2\right)}^{2}} + C .$