# How do you integrate int(x+1)/((2x-4)(x+5)(x-2)) using partial fractions?

First we observe that

$\int \frac{x + 1}{\left(2 x - 4\right) \left(x + 5\right) \left(x - 2\right)} \mathrm{dx} = \frac{1}{2} \cdot \int \frac{x + 1}{\left(x + 5\right) \cdot {\left(x - 2\right)}^{2}} \mathrm{dx}$

Hence we have to find constants A,B,C such as

$\frac{x + 1}{\left(x + 5\right) \cdot {\left(x - 2\right)}^{2}} = \frac{A}{x + 5} + \frac{B}{x - 2} + \frac{C}{x - 2} ^ 2$

We can calculate these constants by giving x three different values

for example $x = 0 , x = - 1 , x = 3$

hence we get

$\frac{x + 1}{\left(x + 5\right) \cdot {\left(x - 2\right)}^{2}} = - \frac{2}{49 \cdot \left(x + 5\right)} + \frac{2}{49 \cdot \left(x - 2\right)} + \frac{3}{14 {\left(x - 2\right)}^{2}}$

Hence now we have that

int(x+1)/((2x-4)(x+5)(x-2))dx=1/2*int (x+1)/[(x+5)*(x-2)^2]dx= 1/2*[int (-2)/(49*(x+5))dx +int 2/(49(x-2))dx+int 3/[(14)(x-2)^2]dx]= 1/2*[-2/49*ln(x+5)+2/49*ln(x-2)-3/14*(1/(x-2))]+c