# How do you integrate 1 / (x(x^4+1)) using partial fractions?

Dec 8, 2017

The answer is $= \ln \left(| x |\right) - \frac{1}{4} \ln \left({x}^{4} + 1\right) + C$

#### Explanation:

Perform the decomposition into partial fractions

$\frac{1}{x \left({x}^{4} + 1\right)} = \frac{A}{x} + \frac{B {x}^{3} + C}{{x}^{4} + 1}$

$= \frac{A \left({x}^{4} + 1\right) + x \left(B {x}^{3} + C\right)}{x \left({x}^{4} + 1\right)}$

The denominators are the same, compare the numerators

$1 = A \left({x}^{4} + 1\right) + x \left(B {x}^{3} + C\right)$

Let $x = 0$, $\implies$, $1 = A$

Coefficients of ${x}^{4}$

$0 = A + B$, $\implies$, $B = - A = - 1$

Coefficients of $x$

$0 = C$

Therefore,

$\frac{1}{x \left({x}^{4} + 1\right)} = \frac{1}{x} + \frac{- {x}^{3}}{{x}^{4} + 1}$

So,

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

$= \ln \left(| x |\right) - \int \frac{{x}^{3} \mathrm{dx}}{{x}^{4} + 1}$

Perform the second part by substitution

Let $u = {x}^{4} + 1$, $\implies$, $\mathrm{du} = 4 {x}^{3} \mathrm{dx}$

So,

$\int \frac{{x}^{3} \mathrm{dx}}{{x}^{4} + 1} = \frac{1}{4} \int \frac{\mathrm{du}}{u}$

$= \frac{1}{4} \ln u$

$= \frac{1}{4} \ln \left({x}^{4} + 1\right)$

Finally,

$\int \frac{\mathrm{dx}}{x \left({x}^{4} + 1\right)} = \ln \left(| x |\right) - \frac{1}{4} \ln \left({x}^{4} + 1\right) + C$

Dec 8, 2017

$\int \frac{\mathrm{dx}}{x \cdot \left({x}^{4} + 1\right)} = L n x - \frac{1}{4} L n \left({x}^{4} + 1\right) + C$

#### Explanation:

$\int \frac{\mathrm{dx}}{x \cdot \left({x}^{4} + 1\right)}$

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

=$\int \frac{\mathrm{dx}}{x}$-$\int \frac{{x}^{3} \cdot \mathrm{dx}}{{x}^{4} + 1}$

=$L n x - \frac{1}{4} \int \frac{4 {x}^{3} \cdot \mathrm{dx}}{{x}^{4} + 1}$

=$L n x - \frac{1}{4} L n \left({x}^{4} + 1\right) + C$