Question

Solve : $\int \frac{x^6-2x^4+3x^3-9x^2+4}{x^5-5x^3+4x}dx$.

Answer 1

Given the integral,

$\int \frac{x^6-2x^4+3x^3-9x^2+4}{x^5-5x^3+4x}dx$

Applying polynomial long division,

$=\int (\frac{3x^4+3x^3-13x^2+4}{x^5-5x^3+4x}+x)dx=\int \frac{3x^4+3x^3-13x^2+4}{x^5-5x^3+4x}dx+\int xdx$

For,

$\int \frac{3x^4+3x^3-13x^2+4}{x^5-5x^3+4x}dx$

Applying factorisation of the denominator we get,

$=\int \frac{3x^4+3x^3-13x^2+4}{(x-2)(x-1)x(x+1)(x+2)}dx$

Again using partial fraction,

$=\int (-\frac{1}{x+2}+\frac{3}{2(x+1)}+\frac{1}{x}+\frac{1}{2(x-1)}+\frac{1}{x-2})dx=-\int \frac{1}{x+2}dx+\frac{3}{2}\int \frac{1}{x+1}dx+\int \frac{1}{x}dx+\frac{1}{2}\int \frac{1}{x-1}dx+\int \frac{1}{x-2}dx=-\ln (x+2)+\frac{3\ln (x+1)}{2}+\ln \left(x\right)+\frac{\ln (x-1)}{2}+\ln (x-2)$

And $\int xdx=\frac{x^2}{2}$

$\therefore \int \frac{3x^4+3x^3-13x^2+4}{x^5-5x^3+4x}dx+\int xdx=-\ln (x+2)+\frac{3\ln (x+1)}{2}+\ln \left(x\right)+\frac{\ln (x-1)}{2}+\ln (x-2)+\frac{x^2}{2}$

Hence,

$\int \frac{x^6-2x^4+3x^3-9x^2+4}{x^5-5x^3+4x}dx=-\ln \left(|x+2|\right)+\frac{3\ln \left(|x+1|\right)}{2}+\ln \left(\left|x\right|\right)+\frac{\ln \left(|x-1|\right)}{2}+\ln \left(|x-2|\right)+\frac{x^2}{2}+C.$