# How do you find int 1/((x+7)(x^2+9))dx using partial fractions?

Dec 2, 2017

$\frac{1}{58} \left(\ln | x + 7 | + \frac{7}{3} {\tan}^{-} 1 \left(\frac{x}{3}\right) - \frac{1}{2} \ln \left({x}^{2} + 9\right)\right)$

#### Explanation:

First we need to do partial fractions. We know that the numerator will be of a degree lower than the denominator, so we can write an equation like this:
$\frac{1}{\left(x + 7\right) \left({x}^{2} + 9\right)} = \frac{A}{x + 7} + \frac{B x + C}{{x}^{2} + 9}$

To do partial fractions, we multiply both sides by the left hand side denominator:
$\frac{1}{\cancel{\left(x + 7\right) \left({x}^{2} + 9\right)}} \cancel{\left(x + 7\right) \left({x}^{2} + 9\right)} = \left(x + 7\right) \left({x}^{2} + 9\right) \left(\frac{A}{\cancel{x + 7}} + \frac{B x + C}{\cancel{{x}^{2} + 9}}\right)$

This gives us:
$1 = A \left({x}^{2} + 9\right) + \left(B x + C\right) \left(x + 7\right)$

$1 = A {x}^{2} + 9 A + B {x}^{2} + 7 B x + C x + 7 C$

If we group coefficients of the same degree, we have:
$1 = \left(A + B\right) {x}^{2} + \left(7 B + C\right) x + \left(9 A + 7 C\right)$

We know how much of each degree of $x$ is on the left, so we can setup the following system of equations:
$A + B = 0$
$7 B + C = 0$
$9 A + 7 C = 1$

Solving it, we get:
$A = \frac{1}{58}$
$B = - \frac{1}{58}$
$C = \frac{7}{58}$

We can now plug this back into our integral to get:
$\int \setminus \frac{\frac{1}{58}}{x + 7} + \frac{- \frac{1}{58} x + \frac{7}{58}}{{x}^{2} + 9} \setminus \mathrm{dx}$

To keep track, I will name the left integral Integral 1 and the right one Integral 2

Integral 1
$\int \setminus \frac{\frac{1}{58}}{x + 7} \setminus \mathrm{dx} = \frac{1}{58} \int \setminus \frac{1}{x + 7} \setminus \mathrm{dx}$

We can solve it by letting $u = x + 7$ and $\frac{\mathrm{du}}{\mathrm{dx}} = 1$:
$\frac{1}{58} \int \setminus \frac{1}{u} \setminus \mathrm{du} = \frac{1}{58} \ln | x + 7 |$

Integral 2
$\int \setminus \frac{- \frac{1}{58} x + \frac{7}{58}}{{x}^{2} + 9} \setminus \mathrm{dx} = \frac{1}{58} \int \setminus \frac{- x + 7}{{x}^{2} + 9} \setminus \mathrm{dx}$

We can split this into two:
$= \frac{1}{58} \left(- \int \setminus \frac{x}{{x}^{2} + 9} + \int \setminus \frac{7}{{x}^{2} + 9} \setminus \mathrm{dx}\right)$

I will call the left one Integral 3, and the right one Integral 4.

Integral 3
Let $u = {x}^{2} + 9$ and $\frac{\mathrm{du}}{\mathrm{dx}} = 2 x$ and divide through by $2 x$:
$- \int \setminus \frac{x}{{x}^{2} + 9} \setminus \mathrm{dx} = - \int \setminus \frac{1}{2 \cancel{x}} \frac{\cancel{x}}{u} \setminus \mathrm{du} = - \frac{1}{2} \int \setminus \frac{1}{u} \setminus \mathrm{du} = - \frac{1}{2} \ln | u |$

Now we resubstitute, and since ${x}^{2} + 9$ is always positive, we can remove the absolute value part:
$= - \frac{1}{2} \ln \left({x}^{2} + 9\right)$

Integral 4
$\int \setminus \frac{7}{{x}^{2} + 9} \setminus \mathrm{dx} = 7 \int \setminus \frac{1}{{x}^{2} + 9} \setminus \mathrm{dx}$

We want to turn this integral into the familiar form:
$\int \setminus \frac{1}{{x}^{2} + 1} \setminus \mathrm{dx} = {\tan}^{-} 1 \left(x\right)$

To do this, we want to factor out a $\frac{1}{9}$ from the bottom, so I will let $x = \sqrt{9} u = 3 u$ and $\frac{\mathrm{dx}}{\mathrm{du}} = 3$. Note that since we took the derivative with respect to $u$, we have to multiply by it instead of dividing:
$7 \int \setminus \frac{1}{{x}^{2} + 9} \setminus \mathrm{dx} = 7 \int \setminus 3 \cdot \frac{1}{{\left(3 u\right)}^{2} + 9} \setminus \mathrm{du} = 21 \int \setminus \frac{1}{9 {u}^{2} + 9} \setminus \mathrm{du}$

Now we can factor out on the bottom:
$21 \int \setminus \frac{1}{9 {u}^{2} + 9} \setminus \mathrm{du} = \frac{21}{9} \int \setminus \frac{1}{{u}^{2} + 1} \setminus \mathrm{du} = \frac{21}{9} {\tan}^{-} 1 \left(u\right)$

Solving for $u$ in $x = 3 u$, we get $u = \frac{x}{3}$. If we resubstitute, we get:
$\frac{21}{9} {\tan}^{-} 1 \left(\frac{x}{3}\right)$

Completing Integral 2
Now we can plug in Integral 3 and 4 to get:
$\frac{1}{58} \int \setminus \frac{- x + 7}{{x}^{2} + 9} \setminus \mathrm{dx} = \frac{1}{58} \left(- \frac{1}{2} \ln \left({x}^{2} + 9\right) + \frac{21}{9} {\tan}^{-} 1 \left(\frac{x}{3}\right)\right)$

Completing the original integral
We know what Integral 1 and 2 is equal to, so we can solve the original integral:
$\int \setminus \frac{\frac{1}{58}}{x + 7} + \frac{- \frac{1}{58} x + \frac{7}{58}}{{x}^{2} + 9} \setminus \mathrm{dx} =$

$= \frac{1}{58} \ln | x + 7 | + \frac{1}{58} \left(- \frac{1}{2} \ln \left({x}^{2} + 9\right) + \frac{21}{9} {\tan}^{-} 1 \left(\frac{x}{3}\right)\right)$

Simplifying, we get:
$= \frac{1}{58} \left(\ln | x + 8 | + \frac{7}{3} {\tan}^{-} 1 \left(\frac{x}{3}\right) - \frac{1}{2} \ln \left({x}^{2} + 9\right)\right)$