How do you find the indefinite integral of $int (x(x+2))/(x^3+3x^2-4)$ ?

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Answer 1 · 2018-06-01T05:39:22

$1/3ln|(x+2)^2(x-1)|+C, OR, ln|(x+2)^(2/3)(x-1)^(1/3)|+C$ .

Let, $I=int(x(x+2))/(x^3+3x^2-4)dx$ .

The sum of the co-effs. of the poly. in the Dr. is $0$ .

$:. (x-1)$ is its factor.

We have, $x^3+3x^2-4=ul(x^3-x^2)+ul(4x^2-4x)+ul(4x-4)$ ,

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

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

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

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

Observe that, $2(x-1)+1(x+2)=3x$ . Therefore,

$I=int(x(x+2))/(x^3+3x^2-4)dx$

$=intx/((x-1)(x+2))dx$ ,

$=1/3int(3x)/((x-1)(x+2))dx$ ,

$=1/3int{2(x-1)+1(x+2)}/((x-1)(x+2))dx$ ,

$=1/3int{(2(x-1))/((x-1)(x+2))+(1(x+2))/((x-1)(x+2))}dx$ ,

$=1/3int{2/(x+2)+1/(x-1)}dx$ ,

$=1/3{2ln|(x+2)|+ln|(x-1)|}$ .

$:.I=1/3ln|(x+2)^2(x-1)|+C, or,$

$I=ln|(x+2)^(2/3)(x-1)^(1/3)|+C$ .

Enjoy Maths.!

How do you find the indefinite integral of int (x(x+2))/(x^3+3x^2-4)int (x(x+2))/(x^3+3x^2-4)?