# How do you integrate ((3x^2)-25x+43)dx/((2x+1)((x-2)^2))?

Sep 14, 2015

$\frac{9}{2} \ln \left\mid 2 x + 1 \right\mid - 3 \ln \left\mid x - 2 \right\mid - \frac{1}{x - 2} + C$

#### Explanation:

Lets rewrite integrand $\frac{3 {x}^{2} - 25 x + 43}{\left(2 x + 1\right) {\left(x - 2\right)}^{2}}$ as follows:

$\frac{3 {x}^{2} - 25 x + 43}{\left(2 x + 1\right) {\left(x - 2\right)}^{2}} = \frac{A}{2 x + 1} + \frac{B}{x - 2} + \frac{C}{x - 2} ^ 2 =$

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

$= \frac{A \left({x}^{2} - 4 x + 4\right) + B \left(2 {x}^{2} - 3 x - 2\right) + C \left(2 x + 1\right)}{\left(2 x + 1\right) {\left(x - 2\right)}^{2}} =$

$= \frac{A {x}^{2} - 4 A x + 4 A + 2 B {x}^{2} - 3 B x - 2 B + 2 C x + C}{\left(2 x + 1\right) {\left(x - 2\right)}^{2}} =$

$= \frac{{x}^{2} \left(A + 2 B\right) + x \left(- 4 A - 3 B + 2 C\right) + \left(4 A - 2 B + C\right)}{\left(2 x + 1\right) {\left(x - 2\right)}^{2}}$

Comparing with integrand coefficients we get system of linear equations with 3 unknowns:

$A + 2 B = 3 \implies A = 3 - 2 B$
$- 4 A - 3 B + 2 C = - 25$
$4 A - 2 B + C = 43$

From 1st eq insert into the 2nd and 3rd:

$- 4 \left(3 - 2 B\right) - 3 B + 2 C = - 25$
$4 \left(3 - 2 B\right) - 2 B + C = 43$

$- 12 + 8 B - 3 B + 2 C = - 25$
$12 - 8 B - 2 B + C = 43$

$5 B + 2 C = - 13$
$- 10 B + C = 31$

$5 B + 2 C = - 13$
$- 20 B + 2 C = 62$

$25 B = - 75 \implies B = - 3$
$C = 31 + 10 B = 31 - 30 = 1 \implies C = 1$
$A = 3 - 2 B = 3 + 6 = 9 \implies A = 9$

So, the integral is broken apart:

$\int \frac{9}{2 x + 1} \mathrm{dx} + \int \frac{- 3}{x - 2} \mathrm{dx} + \int \frac{1}{x - 2} ^ 2 \mathrm{dx} = I$

$2 x + 1 = t , 2 \mathrm{dx} = \mathrm{dt} , \mathrm{dx} = \frac{\mathrm{dt}}{2} ,$

$x - 2 = u , \mathrm{dx} = \mathrm{du} ,$

$x - 2 = m , \mathrm{dx} = \mathrm{dm}$

$I = \int \frac{9}{t} \frac{\mathrm{dt}}{2} + \int \frac{- 3}{u} \mathrm{du} + \int \frac{1}{m} ^ 2 \mathrm{dm} =$

$= \frac{9}{2} \int \frac{\mathrm{dt}}{t} - 3 \int \frac{\mathrm{du}}{u} + \int {m}^{-} 2 \mathrm{dm} =$

$= \frac{9}{2} \ln \left\mid t \right\mid - 3 \ln \left\mid u \right\mid + {m}^{-} \frac{1}{-} 1 + C =$

$= \frac{9}{2} \ln \left\mid 2 x + 1 \right\mid - 3 \ln \left\mid x - 2 \right\mid - \frac{1}{x - 2} + C$