# What is int_(2)^(8) (x-1)/(x^3+x^2)dx ?

Aug 14, 2017

${\int}_{2}^{8} \frac{x - 1}{{x}^{3} + {x}^{2}} = 2 \ln | \frac{4}{3} | - \frac{3}{8}$

#### Explanation:

The denominator can be factored as

${x}^{2} \left(x + 1\right)$

Now using partial fractions, we have:

$\frac{A x + B}{x} ^ 2 + \frac{C}{x + 1} = \frac{x - 1}{x \left(x + 1\right)}$

$A {x}^{2} + B x + A x + B + C {x}^{2} = x - 1$

We now write a system of equation $\left\{\begin{matrix}A + C = 0 \\ A + B = 1 \\ B = - 1\end{matrix}\right.$

If we solve, we get:

$A = 2 , B = - 1 , C = - 2$

Hence the partial fraction decomposition is

$\frac{2 x - 1}{x} ^ 2 - \frac{2}{x + 1}$

The integral becomes:

$I = \int \frac{2 x - 1}{x} ^ 2 - \frac{2}{x + 1} \mathrm{dx}$

$I = \int \frac{2 x - 1}{x} ^ 2 \mathrm{dx} - \int \frac{2}{x + 1} \mathrm{dx}$

$I = \int \frac{2 x}{x} ^ 2 - \frac{1}{x} ^ 2 \mathrm{dx} - \int \frac{2}{x + 1} \mathrm{dx}$

$I = \int \frac{2}{x} \mathrm{dx} - \int \frac{1}{x} ^ 2 \mathrm{dx} - \int \frac{2}{x + 1} \mathrm{dx}$

$I = 2 \ln | x | + {x}^{-} 1 - 2 \ln | x + 1 | + C$

Which can be written as

$I = 2 \ln | \frac{x}{x + 1} | + {x}^{-} 1 + C$

We now evaluate the definite integral.

$I = 2 \ln | \frac{8}{9} | + \frac{1}{8} - \left(2 \ln | \frac{2}{3} | + \frac{1}{2}\right)$

$I = 2 \ln | \frac{8}{9} | - 2 \ln | \frac{2}{3} | + \frac{1}{8} - \frac{1}{2}$

$I = 2 \left(\ln | \frac{\frac{8}{9}}{\frac{2}{3}} |\right) - \frac{3}{8}$

$I = 2 \ln | \frac{4}{3} | - \frac{3}{8}$

Hopefully this helps!