A problem posted by Shafiqur Rahman

Evalute $\int_0^2 \frac{x^4 \; dx}{x^4-4x^3+12x^2-16x+8} $

Answer:

$\frac{x^4}{x^4-4x^3+12x^2-16x+8} $

$= 1 + \frac{4x^3-12x^2+16x-8}{x^4-4x^3+12x^2-16x+8} $

$= 1 + \frac{4x^3-12x^2+24x-16-8x+8}{x^4-4x^3+12x^2-16x+8} $

$= 1 + \frac{4x^3-12x^2+24x-16}{x^4-4x^3+12x^2-16x+8}- \frac{8(x-1)}{x^4-4x^3+12x^2-16x+8} $

So,

$\int \frac{x^4 \; dx}{x^4-4x^3+12x^2-16x+8} $

$= x + ln\left|x^4-4x^3+12x^2-16x+8\right| - 8\int \frac{x-1}{x^4-4x^3+12x^2-16x+8}\; dx $

$x^4-4x^3+12x^2-16x+8 $

$=(x^2 - 2x)^2 + 8(x^2 - 2x) + 8 $

Substituting $u=x^2 - 2x$ and $du=2(x-1)\;dx$, we gat

$8\int \frac{x-1}{x^4-4x^3+12x^2-16x+8}\; dx $

$=4\int \frac{du}{u^2+8u+8} $

$=\int \frac{du}{(u+4)^2 - 8} $

Substituting $w = u+2$ and $dw=du$, we gat

$\int \frac{du}{(u+4)^2 - 8} $


$=\int \frac{dw}{w^2 - 8} $

$= \frac{1}{4\sqrt{2}} \ln\left|\frac{w- 2\sqrt{2}}{w+2\sqrt{2}}\right| $

So, $\frac{8(x-1)}{x^4-4x^3+12x^2-16x+8}$

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

Thus we have,

$\int \frac{x^4}{x^4-4x^3+12x^2-16x+8} $

$= x + ln\left|x^4-4x^3+12x^2-16x+8\right| -\frac{1}{\sqrt{2}} \ln\left| \frac{x^2-2x+4 -2\sqrt{2}}{x^2-2x+4 +2\sqrt{2}}\right| + C $

And,

$\int_0^2 \frac{x^4}{x^4-4x^3+12x^2-16x+8} $

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

- \left(0 + \ln 2 -\frac{1}{\sqrt{2}} \ln\left| \frac{4 -2\sqrt{2}}{4 +2\sqrt{2}}\right|\right) $

$=2$