# How do you integrate (x^3+25)/(x^2+4x+3) using partial fractions?

Jul 8, 2017

$12 \ln | x + 1 | + \ln | x + 3 | + c$

#### Explanation:

$\text{factorising the numerator}$

(x^3+25)/((x+1)(x+3)

$\Rightarrow \frac{{x}^{3} + 25}{\left(x + 1\right) \left(x + 3\right)} = \frac{A}{x + 1} + \frac{B}{x + 3}$

$\text{multiply through by } \left(x + 1\right) \left(x + 3\right)$

$\Rightarrow {x}^{3} + 25 = A \left(x + 3\right) + B \left(x + 1\right)$

$\text{using the "color(blue)"cover up method}$

$x = - 3 \to - 2 = - 2 B \Rightarrow B = 1$

$x = - 1 \to 24 = 2 A \Rightarrow A = 12$

$\Rightarrow \int \frac{{x}^{3} + 25}{{x}^{2} + 4 x + 3} \mathrm{dx} = \int \frac{12}{x + 1} \mathrm{dx} + \int \frac{1}{x + 3} \mathrm{dx}$

$= 12 \ln | x + 1 | + \ln | x + 3 | + c$

Jul 9, 2017

$\int \frac{{x}^{3} + 25}{{x}^{2} + 4 x + 3} \mathrm{dx}$

$= \int \left(x - 4\right) \mathrm{dx} + \int \frac{13 x + 27}{{x}^{2} + 4 x + 3} \mathrm{dx}$

$= {x}^{2} / 2 - 4 x + \int \frac{13 x + 27}{\left(x + 1\right) \cdot \left(x + 3\right)} \mathrm{dx}$

Now I decomposed $\frac{13 x + 27}{\left(x + 1\right) \cdot \left(x + 3\right)} = \frac{A}{x + 1} + \frac{B}{x + 3}$ fraction

$A \cdot \left(x + 3\right) + B \cdot \left(x + 1\right) = 13 x + 27$

$\left(A + B\right) \cdot x + 3 A + B = 13 x + 27$

After equating coefficients, I found $A + B = 13$ and $3 A + B = 27$ equations,

After solving them simultaneously, $A = 7$ and $B = 6$

Thus,

$\int \frac{{x}^{3} + 25}{{x}^{2} + 4 x + 3} \mathrm{dx}$

$= {x}^{2} / 2 - 4 x + \int \frac{13 x + 27}{\left(x + 1\right) \cdot \left(x + 3\right)} \mathrm{dx}$

$= {x}^{2} / 2 - 4 x + \int \frac{7}{x + 1} \mathrm{dx} + \int \frac{6}{x + 3} \mathrm{dx}$

$= {x}^{2} / 2 - 4 x + 7 L n \left(x + 1\right) + 6 L n \left(x + 3\right) + C$

#### Explanation:

1) I took long division

2) I decomposed second integral into basic fractions