Question

How to integrate with steps:

(4x2-6)/(x+5)(x-2)(3x-1)

Answer 1

$\displaystyle\int\frac{4x^2-6}{(x+5)(x-2)(3x-1)}\,\mathrm dx$

You have a rational expression whose numerator's degree is smaller than the denominator's. This tells you you should consider a partial fraction decomposition. We want to rewrite the integrand in the form

$\dfrac{4x^2-6}{(x+5)(x-2)(3x-1)}=\dfrac a{x+5}+\dfrac b{x-2}+\dfrac c{3x-1}$

$\implies4x^2-6=a(x-2)(3x-1)+b(x+5)(3x-1)+c(x+5)(x-2)$

You can use the "cover-up" method here to easily solve for $a,b,c$. It involves fixing a value of $x$ to make 2 of the 3 terms on the right side disappear and leaving a simple algebraic equation to solve for the remaining one.

• If $x=-5$, then $94=112a\implies a=\dfrac{47}{56}$
• If $x=2$, then $10=35b\implies b=\dfrac27$
• If $x=\dfrac13$, then $-\dfrac{50}9=-\dfrac{80}9c\implies c=\dfrac58$

So the integral we want to compute is the same as

$\displaystyle\frac{47}{56}\int\frac{\mathrm dx}{x+5}+\frac{10}{35}\int\frac{\mathrm dx}{x-2}+\frac58\int\frac{\mathrm dx}{3x-1}$

and each integral here is trivial. We end up with

$\displaystyle\int\frac{4x^2-6}{(x+5)(x-2)(3x-1)}\,\mathrm dx=\frac{47}{56}\ln|x+5|+\frac27\ln|x-2|+\frac5{24}\ln|3x-1|+C$

which can be condensed as

$\ln\left|(x+5)^{47/56}(x-2)^{2/7}(3x-1)^{5/24}\right|+C$