# How do you use partial fraction decomposition to integrate (−10x^2+27x−14)/((x−1)^3(x+2))?

Oct 21, 2017

int(−10x^2+27x−14)/((x−1)^3(x+2))dx=
$\int \left(\frac{4}{x + 2} - \frac{4}{x - 1} + \frac{2}{x - 1} ^ 2 + \frac{1}{x - 1} ^ 3\right) \mathrm{dx}$

#### Explanation:

Since the question is asking about how to use partial fractions to perform the integration, I shall only provide the partial fraction decomposition and leave the rest to the questioner.

(−10x^2+27x−14)/((x−1)^3(x+2))=
$\frac{A}{x + 2} + \frac{B}{x - 1} + \frac{C}{x - 1} ^ 2 + \frac{D}{x - 1} ^ 3$

−10x^2+27x−14=
$A {\left(x - 1\right)}^{3} + B {\left(x - 1\right)}^{2} \left(x + 2\right) + C \left(x - 1\right) \left(x + 2\right) + D \left(x + 2\right)$

Let $x = 2$

Then $3 = 3 D \Rightarrow D = 1$

Let $x = - 2$

Then $- 108 = - 27 A \Rightarrow A = 4$

Notice that if we had expanded the RHS, we would've found two cubic terms with coefficients $A$ and $B$. And since $L H S = R H S , \left(A + B\right) {x}^{3} = {x}^{3} \Rightarrow A + B = 0 \Rightarrow 4 + B = 0 \Rightarrow B = - 4$

Finally letting $x = 0$, we get

$- 14 = A + 2 B - 2 C + 2 D$

By simplifying and subbing in the values we already have, we get $C = 2$

therefore(−10x^2+27x−14)/((x−1)^3(x+2))=
$\frac{4}{x + 2} - \frac{4}{x - 1} + \frac{2}{x - 1} ^ 2 + \frac{1}{x - 1} ^ 3$

Thus int(−10x^2+27x−14)/((x−1)^3(x+2))dx=
$\int \left(\frac{4}{x + 2} - \frac{4}{x - 1} + \frac{2}{x - 1} ^ 2 + \frac{1}{x - 1} ^ 3\right) \mathrm{dx}$