# How do you integrate (3x^2+9x-4)/((x-1)(x^2+4x-1)) using partial fractions?

May 20, 2017

#### Explanation:

Find the partial fractions.

$\frac{3 {x}^{2} + 9 x - 4}{\left(x - 1\right) \left({x}^{2} + 4 x - 1\right)} = \frac{A}{x - 1} + \frac{B x + C}{{x}^{2} + 4 x - 1}$

$\left(3 {x}^{2} + 9 x - 4\right) = A \left({x}^{2} + 4 x - 1\right) + \left(B x + C\right) \left(x - 1\right)$

Eliminate B and C by letting x = 1:

$\left(3 {\left(1\right)}^{2} + 9 \left(1\right) - 4\right) = A \left({\left(1\right)}^{2} + 4 \left(1\right) - 1\right)$

$8 = A 4$

$A = 2$

$\left(3 {x}^{2} + 9 x - 4\right) = 2 \left({x}^{2} + 4 x - 1\right) + \left(B x + C\right) \left(x - 1\right)$

Eliminate B by letting x = 0:

$\left(3 {\left(0\right)}^{2} + 9 \left(0\right) - 4\right) = 2 \left({\left(0\right)}^{2} + 4 \left(0\right) - 1\right) + C \left(\left(0\right) - 1\right)$

$- 4 = - 2 - C$

$C = 2$

$\left(3 {x}^{2} + 9 x - 4\right) = 2 \left({x}^{2} + 4 x - 1\right) + \left(B x + 2\right) \left(x - 1\right)$

Let x = -1:

$\left(3 {\left(- 1\right)}^{2} + 9 \left(- 1\right) - 4\right) = 2 \left({\left(- 1\right)}^{2} + 4 \left(- 1\right) - 1\right) + \left(B \left(- 1\right) + 2\right) \left(- 1 - 1\right)$

$- 10 = - 8 + 2 B - 4$

$B = 1$

$\int \frac{3 {x}^{2} + 9 x - 4}{\left(x - 1\right) \left({x}^{2} + 4 x - 1\right)} \mathrm{dx} = 2 \int \frac{1}{x - 1} \mathrm{dx} + \int \frac{x + 2}{{x}^{2} + 4 x - 1} \mathrm{dx}$

Multiply the second integral by 1/2 so that the numerator can be multiplied by 2:

$\int \frac{3 {x}^{2} + 9 x - 4}{\left(x - 1\right) \left({x}^{2} + 4 x - 1\right)} \mathrm{dx} = 2 \int \frac{1}{x - 1} \mathrm{dx} + \frac{1}{2} \int \frac{2 x + 4}{{x}^{2} + 4 x - 1} \mathrm{dx}$

Both integrals become the natural logarithm:

$\int \frac{3 {x}^{2} + 9 x - 4}{\left(x - 1\right) \left({x}^{2} + 4 x - 1\right)} \mathrm{dx} = 2 \ln | x - 1 | + \frac{1}{2} \ln | {x}^{2} + 4 x - 1 | + C$