# How do you integrate int (3x+7)/(x^4-16)dx using partial fractions?

Dec 6, 2015

$\frac{1}{32} \left[13 \ln | x - 2 | - \ln | x + 2 | - 6 \ln \left({x}^{2} + 4\right) - 14 \arctan \left(\frac{x}{2}\right)\right] + C$

#### Explanation:

*Step 1:* Partial fraction decomposition
We begin be factor the denominator

$\frac{3 x + 7}{{x}^{4} - 16} = \frac{3 x + 7}{\left({x}^{2} + 4\right) \left({x}^{2} - 4\right)} = \frac{3 x + 7}{\left(x - 2\right) \left(x + 2\right) \left({x}^{2} + 4\right)}$

$= \frac{A}{x - 2} + \frac{B}{x - 2} + \frac{C x + D}{{x}^{2} + 4} \text{ " " " " " " } \textcolor{red}{\left(1\right)}$

$3 x + 7 = A \left(x + 2\right) \left({x}^{2} + 4\right) + B \left(x - 2\right) \left({x}^{2} + 4\right) \text{ " " " " " " " + (Cx+D)(x^2-4)" " " " " "color(red)((2) }$

FOIL the partial fraction expression we have
$A \left({x}^{3} + 2 {x}^{2} + 4 x + 8\right)$
$B \left({x}^{3} - 2 {x}^{2} + 4 x - 8\right)$
$C {x}^{3} + D {x}^{2} - 4 C x - 4 D$

$0 {x}^{3} : \text{ " A+B + C = 0 " " " " } \left(E q . 3\right)$
$0 {x}^{2} : \text{ " 2A-2B+D= 0 " " " " } \left(E q . 4\right)$
$3 x : \text{ " "4A+4B-4C= 3" " " " " " } \left(E q . 5\right)$
$7 : \text{ " " 8 A-8B -4D = 7" " " " " " } \left(E q . 6\right)$

Step 2: Solve the system of equation above

Multiply 4 to $E q .3$ and add to $E q .5$
$4 \cdot \left(A + B + C = 0\right) \Leftrightarrow 4 A + 4 B + 4 C = 0$

$+ \left(4 A + 4 B - 4 C = 3\right)$
$\textcolor{red}{8 A + 8 B = 3} \text{ " } \left(E q . 7\right)$

Multiply $\left(E q . 4\right)$ by 4 and add to $\left(E q .6\right)$
$4 \left(2 A - 2 B + D = 0\right) \Leftrightarrow 8 A - 8 B + 4 D = 0$

+$8 A - 8 B + 4 D = 7$
$\textcolor{red}{16 A - 16 B = 7} \left(E q . 8\right)$

Multiply $\left(E q .7\right)$ by 2, add to $\left(E q .8\right)$
$2 \left(8 a + 8 b = 3\right) \Leftrightarrow 16 A - 16 B = 6$

$+ 16 A - 16 B = 7$
$32 A = 13 \Leftrightarrow A = \frac{13}{32} \text{ " " } \left(8\right)$

Substitute into either $\left(E q .7\right)$ or $\left(E q .8\right)$ to solve for B
$\frac{13}{32} + 8 B = 3 \Leftrightarrow B = - \frac{1}{32} \text{ " " " " } \left(9\right)$

Substitute (8) and (9) into Eq. 3 and Eq.4 to solve for C and D
$C = - \frac{3}{8} \text{ " " " } \left(10\right)$
$D = - \frac{7}{8} \text{ " " " } \left(11\right)$

Step 3: Rewrite the partial faction using (1) and (8) (9) (10)(11)
(3x+7)/(x^4-16) = 13/(32(x-2)) -(1/(32(x+2))) - (3x+7)/(8(x^2 +4)

Step 4: Rewrite the integral
$\int \left(\frac{3 x + 7}{{x}^{4} - 16}\right) \mathrm{dx} = \frac{13}{32} \int \left(\frac{1}{x - 2}\right) \mathrm{dx} - \frac{1}{32} \int \left(\frac{1}{x + 2}\right) \mathrm{dx} - \frac{3}{8} \cdot \int \left(\frac{x}{{x}^{2} + 4}\right) \mathrm{dx} - \frac{7}{8} \int \left(\frac{1}{{x}^{2} + 4}\right) \mathrm{dx}$

Step 5: Start integrating
$= \frac{13}{32} \ln | x - 2 | - \frac{1}{32} \ln | x + 2 | - \frac{3}{16} \ln \left({x}^{2} + 4\right) - \frac{7}{8} \arctan \left(\frac{x}{2}\right) + C$

$\frac{1}{32} \left[13 \ln | x - 2 | - \ln | x + 2 | - 6 \ln \left({x}^{2} + 4\right) - 14 \arctan \left(\frac{x}{2}\right)\right] + C$

Sorry, this problem is really really long....

Please feel free to edit this, if you see any mistake.