# How do you find int (x-1) / (x(x^4+1)) dx using partial fractions?

Oct 31, 2017

The answer is $= \frac{1}{4} \ln \left(| {x}^{4} + 1 |\right) - \ln \left(| x |\right) + \frac{1}{4 \sqrt{2}} \left(- \ln | 2 {x}^{2} - 2 \sqrt{2} x + 2 |\right) + 2 \arctan \left(2 x - 1\right) + \frac{1}{4 \sqrt{2}} \left(\ln | 2 {x}^{2} + 2 \sqrt{2} x + 2 |\right) + 2 \arctan \left(2 x + 1\right) + C$

#### Explanation:

Perform the decomposition into partial fractions

$\frac{x - 1}{x \left({x}^{4} + 1\right)} = \frac{A {x}^{3} + B {x}^{2} + C}{{x}^{4} + 1} + \frac{D}{x}$

$= \frac{\left(A {x}^{3} + B {x}^{2} + C\right) x + D \left({x}^{4} + 1\right)}{x \left({x}^{4} + 1\right)}$

The denominators are the same, compare the numerators

$x - 1 = A {x}^{4} + B {x}^{3} + C x + D {x}^{4} + D$

Therefore, comparing the LHS and the RHS

$A + D = 0$

$C = 1$

$D = - 1$

$B = 0$

$A = - D = 1$

So,

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

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

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

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

Therefore,

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

$\int \frac{\mathrm{dx}}{x} = \ln \left(| x |\right)$

$\frac{1}{{x}^{4} + 1} = \frac{1}{\left({x}^{2} - \sqrt{2} x + 1\right) \left({x}^{2} + \sqrt{2} x + 1\right)}$

$= \frac{A x + B}{{x}^{2} - \sqrt{2} x + 1} + \frac{C x + D}{{x}^{2} + \sqrt{2} x + 1}$

$= \frac{\left(A x + B\right) \left({x}^{2} + \sqrt{2} x + 1\right) + \left(C x + D\right) \left({x}^{2} - \sqrt{2} x + 1\right)}{\left({x}^{2} - \sqrt{2} x + 1\right) \left({x}^{2} + \sqrt{2} x + 1\right)}$

The denominators are the same, we compare the numerators

$1 = \left(A x + B\right) \left({x}^{2} + \sqrt{2} x + 1\right) + \left(C x + D\right) \left({x}^{2} - \sqrt{2} x + 1\right)$

Let $x = 0$, $\implies$, $1 = B + D$

Coefficients of ${x}^{3}$, $\implies$, $A + C = 0$

Coefficients of $x$, $\implies$, $\sqrt{2} B + A + C - \sqrt{2} D = 0$

$\implies$, $B = D$, $\implies$,$B = D = \frac{1}{2}$

Coefficients of ${x}^{2}$, $\implies$, $A \sqrt{2} + B - \sqrt{2} C + D = 0$

$\implies$, $A = - \frac{1}{2 \sqrt{2}}$

$\implies$, $C = \frac{1}{2 \sqrt{2}}$

So,

$\frac{1}{{x}^{4} + 1} = \frac{- \frac{1}{2 \sqrt{2}} x + \frac{1}{2}}{{x}^{2} - \sqrt{2} x + 1} + \frac{\frac{1}{2 \sqrt{2}} x + \frac{1}{2}}{{x}^{2} + \sqrt{2} x + 1}$

$= \frac{\sqrt{2} - x}{2 \sqrt{2} \left({x}^{2} - \sqrt{2} x + 1\right)} + \frac{\sqrt{2} + x}{2 \sqrt{2} \left({x}^{2} + \sqrt{2} x + 1\right)}$

$\int \frac{\mathrm{dx}}{{x}^{4} + 1} = \int \frac{\left(\sqrt{2} - x\right) \mathrm{dx}}{2 \sqrt{2} \left({x}^{2} - \sqrt{2} x + 1\right)} + \int \frac{\left(\sqrt{2} + x\right) \mathrm{dx}}{2 \sqrt{2} \left({x}^{2} + \sqrt{2} x + 1\right)}$

$\int \frac{\left(\sqrt{2} - x\right) \mathrm{dx}}{2 \sqrt{2} \left({x}^{2} - \sqrt{2} x + 1\right)} = \frac{1}{4 \sqrt{2}} \left(- \ln | 2 {x}^{2} - 2 \sqrt{2} x + 2 |\right) + 2 \arctan \left(2 x - 1\right)$

$\int \frac{\left(\sqrt{2} + x\right) \mathrm{dx}}{2 \sqrt{2} \left({x}^{2} + \sqrt{2} x + 1\right)} = \frac{1}{4 \sqrt{2}} \left(\ln | 2 {x}^{2} + 2 \sqrt{2} x + 2 |\right) + 2 \arctan \left(2 x + 1\right)$