How do you integrate $\frac{x^{3} + 2}{x^{2} - x}$ $(x^3+2)/(x^2-x)$ using partial fractions?

Question

How do you integrate $\frac{x^{3} + 2}{x^{2} - x}$ $(x^3+2)/(x^2-x)$ using partial fractions?

1 Answer

Impact of this question

1347 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-18T14:04:44

$= \frac{x^{2}}{2} + x - 2 ln x + 3 ln (x - 1) + C$ $=x^2/2+x-2lnx+3ln(x-1)+C$

Explanation:

First make a long division to determine
$\frac{x^{3} + 2}{x^{2} - x} = x + 1 + \frac{x + 2}{x^{2} - x}$ $(x^3+2)/(x^2-x)=x+1+(x+2)/(x^2-x)$
To simplify the fraction we use partial fraction
$\frac{x + 2}{x^{2} - x} = \frac{x + 2}{x (x - 1)} = \frac{A}{x} + \frac{B}{x - 1} = \frac{A (x - 1) + B x}{x (x - 1)}$ $(x+2)/(x^2-x)=(x+2)/(x(x-1))=A/x+B/(x-1)=(A(x-1)+Bx)/(x(x-1))$
So we determine A and B
$x + 2 = A (x - 1) + B x$ $x+2=A(x-1)+Bx$
Cmparing the coefficients, we find
$1 = A + B$ $1=A+B$ and $2 = - A$ $2=-A$ and we conclude $B = 3$ $B=3$
$\int \frac{(x^{3} + 2) d x}{x^{2} - x} = \int (x + 1) d x - \int 2 \frac{d x}{x} + \int 3 \frac{d x}{x - 1}$ $int((x^3+2)dx)/(x^2-x)=int(x+1)dx-int2dx/x+int3dx/(x-1)$
$= \frac{x^{2}}{2} + x - 2 ln x + 3 ln (x - 1) + C$ $=x^2/2+x-2lnx+3ln(x-1)+C$

Science

Math

Humanities

... and beyond

How do you integrate $\frac{x^{3} + 2}{x^{2} - x}$ $(x^3+2)/(x^2-x)$ using partial fractions?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you integrate x3+2x2−x(x^3+2)/(x^2-x) using partial fractions?

1 Answer

Explanation:

Related questions

Impact of this question

How do you integrate $\frac{x^{3} + 2}{x^{2} - x}$ $(x^3+2)/(x^2-x)$ using partial fractions?