MHB Solve Int Partial Fractions: $\int\frac{6{x}^{2}+22x-23} {(2x-1)(x+3)(x-2)} dx$

[karush]

$\int\frac{6{x}^{2}+22x-23} {(2x-1)(x+3)(x-2)} dx $

Solve using partial fractions

$\frac{A}{2x-1}+\frac{B}{x+3}-\frac{C}{x-2}$

I pursued got A=2 B=-1 C=-3

Then?

 

[Euge]

Hi karush,

What work have you done so far? Please show us what you have and we'll work from there.

 

[karush]

$2\int\frac{1}{2x-1}dx - \int\frac{1 }{x+3}dx+3\int\frac{1}{x-2}dx$

 

[Euge]

You're right so far. :) Now consider using integral substitution together with the fact $\int du/u = \log|u| + C$ to solve the three integrals you have.

 

[karush]

$2\ln\left({\left| 2x-1 \right|}\right)
-\ln\left({\left|x+3 \right|}\right)
+3\ln\left({\left| x-2 \right|}\right)+C$

the TI gave

$\ln\left({\frac{{\left| x-2 \right|}^{3}\cdot\left| 2x-1 \right|}{\left| x+3 \right|}}\right)+C$

?

 

[Euge]

Your answer is incorrect, but you have just a minor mistake: the coefficient $2$ in $2\ln|2x - 1|$ should not be there. When integrating $2 \int dx/(2x - 1)$, we consider the substitution $u = 2x - 1$. Then $du = 2\, dx$, so $2\int dx/(2x - 1) = \int du/u = \ln|u| + C = \ln|2x - 1| + C$.

 

[karush]

so that's how it disappeared

thanks for help. (Smile)