Answer to Question #124130 in Calculus for Kofi

Question

Evaluate the integeral
∫((2x³ -4x -8)/ (x⁴ -x³ +4x² -4x)) dx

Answer 1

$I = \int\frac{2x^3-4x-8}{x^4-x^3+4x^2-4x}dx.$

This is a fractional-rational function $\frac{2x^3-4x-8}{x^4-x^3+4x^2-4x}$. Factorize the denominator:

$x^4-x^3+4x^2-4x = x^3(x-1)+4x(x-1)=$

$=(x-1)(x^3+4x)=x(x-1)(x^2+4).$

So, we have $\frac{2x^3-4x-8}{x^4-x^3+4x^2-4x}=\frac{2x^3-4x-8}{x(x-1)(x^2+4)}$.

Let's build a decomposition:

$\frac{2x^3-4x-8}{x(x-1)(x^2+4)}=\frac{A}{x}+\frac{B}{x-1}+\frac{Cx+D}{x^2+4}$.

Let's bring the right part to the common denominator:

$\frac{}{}$$\frac{A(x-1)(x^2+4)+Bx(x^2+4)+(Cx+D)(x-1)x}{x(x-1)(x^2+4)}$.

Open the brackets in the numerator:

$A(x^3+4x-x^2-4)+Bx^3+4Bx+Cx^3-$

$-Cx^2+Dx^2-Dx$.

Simplify the expression:

$x^3(A+B+C)+x^2(-A-C+D)+$

$+x(4A+4B-D)+x^0(-4A)$ .

So, we have:

$2x^3-4x-8=x^3(A+B+C)+$

$+x^2(-A-C+D)+x(4A+4B-D)+$

$+x^0(-4A)$.

Let's collect summands with the same degrees:

$A+B+C=2$ ,

$D-A-C=0$ ,

$4A+4B-D=-4$ ,

$-4A=-8$.

From the last equation we find $A = 2$.

Now, we have:

$B+C=0$,

$D-C=2$,

$4B-D=-12$.

Further:

$B=-С$,

$D=2+C$,

$4B-2+B=-12$ .

So, from the last equation we find $B=-2$. And therefore $C=2, D=4$.

Eventually:

$I = \int\frac{2x^3-4x-8}{x^4-x^3+4x^2-4x}dx =$

$=\int\frac{A}{x}+\frac{B}{x-1}+\frac{Cx+D}{x^2+4}=\int\frac{2}{x}dx+\int\frac{-2}{x-1}dx+$

$+\int\frac{2x+4}{x^2+4}dx$.

Let's count each integral separately:

$\int\frac{2}{x}dx = 2ln(x) +C_1$.

$\int\frac{-2}{x-1}dx = -2ln(x-1)+C_2$.

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

$+2arctan(\frac{x}{2}) +C_4=ln(x^2+4)+C_3+$

$2arctan(\frac{x}{2})+C_4$.

Answer:

$I=2ln(x)-2ln(x-1)+ln(x^2+4)+$

$+2arctan(\frac{x}{2})+C$.