Answer to Question #124053 in Calculus for Nii Laryea

Question

Evaluate the integral ∫ (2xᶟ- 4x- 8) / (x⁴- xᶟ + 4x²- 4x) dx

Answer 1

$\displaystyle\int \frac{2x^{3}-4x-8}{x^{4}-x^{3}+4 x^{2}-4x} dx=2\int \frac{x^{3}-2 x-4}{x(x-1)\left(x^{2}+4\right)} dx$

Using partial fraction for the term inside the integral,,

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

Solving for A, B, C and D we get, A = 1, B = -1, C = 1, D = 2.

Thus,

$\displaystyle\int \frac{2x^{3}- 4x -8}{x^{4}-x^{3}+4 x^{2}- 4x} dx =2 \int\left(\frac{1}{x}-\frac{1}{x-1}+\frac{x+2}{x^{2}+4}\right) dx \\~\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=2(\ln x-\ln (x-1))+\int \frac{2 x}{x^{2}+4} dx+2 \int \frac{2}{x^{2}+4} dx\\~\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=2 \ln x-2 \ln (x-1)+\ln \left(x^{2}+4\right)+2 \arctan \left(\frac{x}{2}\right)+c\\~\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\ln x^{2}-\ln (x-1)^{2}+\ln \left(x^{2}+4\right)+2 \arctan \left(\frac{x}{2}\right) +c\\~\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=\ln \left(\frac{x^{2}\left(x^{2}+4\right)}{(x-1)^{2}}\right)+2 \arctan \left(\frac{x}{2}\right)+c$