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

Oct 2, 2017

$\int \frac{\mathrm{dx}}{\left(x + 7\right) \left({x}^{2} + 9\right)} = \frac{1}{58} \ln \left(\frac{\left\mid x + 7 \right\mid}{\sqrt{{x}^{2} + 9}}\right) + \frac{7}{174} \arctan \left(\frac{x}{3}\right) + C$

#### Explanation:

Since one of the factors of the denominator is of second degree we apply partial fractions decomposition as:

$\frac{1}{\left(x + 7\right) \left({x}^{2} + 9\right)} = \frac{A}{x + 7} + \frac{B x + C}{{x}^{2} + 9}$

$\frac{1}{\left(x + 7\right) \left({x}^{2} + 9\right)} = \frac{A \left({x}^{2} + 9\right) + \left(B x + C\right) \left(x + 7\right)}{\left(x + 7\right) \left({x}^{2} + 9\right)}$

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

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

Equation the coefficients of the same degree in $x$:

$\left\{\begin{matrix}A + B = 0 \\ 7 B + C = 0 \\ 9 A + 7 C = 1\end{matrix}\right.$

$\left\{\begin{matrix}A = - B \\ 7 B + C = 0 \\ - 9 B + 7 C = 1\end{matrix}\right.$

$\left\{\begin{matrix}A = - B \\ B = - \frac{C}{7} \\ \frac{9}{7} C + 7 C = 1\end{matrix}\right.$

$\left\{\begin{matrix}A = \frac{1}{58} \\ B = - \frac{1}{58} \\ C = \frac{7}{58}\end{matrix}\right.$

So:

$\int \frac{\mathrm{dx}}{\left(x + 7\right) \left({x}^{2} + 9\right)} = \frac{1}{58} \int \frac{\mathrm{dx}}{x + 7} - \frac{1}{58} \int \frac{x}{{x}^{2} + 9} \mathrm{dx} + \frac{7}{58} \int \frac{\mathrm{dx}}{{x}^{2} + 9}$

Solve the three integrals separately:

$\int \frac{\mathrm{dx}}{x + 7} = \ln \left\mid x + 7 \right\mid + {C}_{1}$

$\int \frac{x}{{x}^{2} + 9} \mathrm{dx} = \frac{1}{2} \int \frac{d \left({x}^{2} + 9\right)}{{x}^{2} + 9} = \frac{1}{2} \ln \left({x}^{2} + 9\right) + {C}_{2}$

$\int \frac{\mathrm{dx}}{{x}^{2} + 9} = \frac{1}{3} \int \frac{d \left(\frac{x}{3}\right)}{{\left(\frac{x}{3}\right)}^{2} + 1} = \frac{1}{3} \arctan \left(\frac{x}{3}\right) + {C}_{3}$

putting the solutions together:

$\int \frac{\mathrm{dx}}{\left(x + 7\right) \left({x}^{2} + 9\right)} = \frac{1}{58} \left(\ln \left\mid x + 7 \right\mid - \frac{1}{2} \ln \left({x}^{2} + 9\right)\right) + \frac{7}{174} \arctan \left(\frac{x}{3}\right) + C$

and finally using the properties of logarithms:

$\int \frac{\mathrm{dx}}{\left(x + 7\right) \left({x}^{2} + 9\right)} = \frac{1}{58} \ln \left(\frac{\left\mid x + 7 \right\mid}{\sqrt{{x}^{2} + 9}}\right) + \frac{7}{174} \arctan \left(\frac{x}{3}\right) + C$