# How do you find int (x+1)/(9x^2+6x+5)dx using partial fractions?

May 13, 2016

$\int \frac{x + 1}{9 {x}^{2} + 6 x + 5} \mathrm{dx}$

= $\frac{1}{18} \ln \left(9 {x}^{2} + 6 x + 5\right) + \frac{1}{9} {\tan}^{- 1} \left(\frac{3 x + 1}{2}\right) + c$

#### Explanation:

As the discriminant of $9 {x}^{2} + 6 x + 5$ is ${6}^{2} - 4 \times 9 \times 5 = 36 - 180 = - 144$, it being negative, cannot be factorized in rational factors.

Observing that differential of $9 {x}^{2} + 6 x + 5$ is $18 x + 6$, let us split $\frac{x + 1}{9 {x}^{2} + 6 x + 5}$ as $\frac{18 x + 18}{18 \left(9 {x}^{2} + 6 x + 5\right)}$ or

$\frac{18 x + 6}{18 \left(9 {x}^{2} + 6 x + 5\right)} + \frac{12}{18 \left(9 {x}^{2} + 6 x + 5\right)}$

Hence $\int \frac{x + 1}{9 {x}^{2} + 6 x + 5} \mathrm{dx}$

= $\int \frac{18 x + 6}{18 \left(9 {x}^{2} + 6 x + 5\right)} \mathrm{dx} + \int \frac{12}{18 \left(9 {x}^{2} + 6 x + 5\right)} \mathrm{dx}$

= $\frac{1}{18} \int \frac{18 x + 6}{9 {x}^{2} + 6 x + 5} \mathrm{dx} + \frac{2}{3} \int \frac{1}{9 {x}^{2} + 6 x + 5} \mathrm{dx}$

Let us integrate first part and assume $u = 9 {x}^{2} + 6 x + 5$, then $\mathrm{du} = \left(18 x + 6\right) \mathrm{dx}$ and it can be written as $\frac{1}{18} \int \frac{\mathrm{du}}{u} = \frac{1}{18} \ln u = \frac{1}{18} \ln \left(9 {x}^{2} + 6 x + 5\right)$.

For second part $\frac{2}{3} \int \frac{1}{9 {x}^{2} + 6 x + 5} \mathrm{dx}$, we will use the identity $\int \frac{\mathrm{dx}}{{x}^{2} + {a}^{2}} = \frac{1}{a} {\tan}^{- 1} \left(\frac{x}{a}\right) + c$. For this converting denominator into sum of two squares, we get $\left(9 {x}^{2} + 6 x + 1\right) + {2}^{2}$, hence

$\frac{2}{3} \int \frac{1}{9 {x}^{2} + 6 x + 5} \mathrm{dx} = \frac{2}{27} \int \frac{1}{\left({x}^{2} + \frac{2}{3} x + \frac{1}{9}\right) + {\left(\frac{2}{3}\right)}^{2}} \mathrm{dx} = \frac{2}{27} \int \frac{1}{{\left(x + \frac{1}{3}\right)}^{2} + {\left(\frac{2}{3}\right)}^{2}} \mathrm{dx}$

Let us now substitute $u = x + \frac{1}{3}$ and as $\mathrm{du} = \mathrm{dx}$ this is now

$\frac{2}{27} \int \frac{1}{{u}^{2} + {\left(\frac{2}{3}\right)}^{2}} \mathrm{du} = \frac{2}{27} \times \left(\frac{3}{2} {\tan}^{- 1} \left(\frac{3 u}{2}\right)\right)$ or

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

Hence, $\int \frac{x + 1}{9 {x}^{2} + 6 x + 5} \mathrm{dx}$

= $\frac{1}{18} \ln \left(9 {x}^{2} + 6 x + 5\right) + \frac{1}{9} {\tan}^{- 1} \left(\frac{3 x + 1}{2}\right) + c$