Question

How to integrate with steps:

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

Answer 1

`\displaystyle\int(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

`(4x^2-6)/((x+5)(x-2)(3x-1))=\frac a{x+5}+\frac b{x-2}+\frac 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=(47)/(56)`
• If
  `x=2`, then
  `10=35b\implies b=\frac27`
• If
  `x=\frac13`, then
  `-\frac{50}9=-\frac{80}9c\implies c=\frac58`

So the integral we want to compute is the same as

`\displaystyle(47)/(56)\int(\mathrm dx)/(x+5)+(10)/(35)\int(\mathrm dx)/(x-2)+\frac58\int(\mathrm dx)/(3x-1)`

and each integral here is trivial. We end up with

`\displaystyle\int(4x^2-6)/((x+5)(x-2)(3x-1))\,\mathrm dx=(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`