Question

What is the integral of `(x^2 +2x-1)/(x^2+9)`?

Answer 1

`=x+ln(x^2+9)-10/3tan^-1(x/3)+C`

Explanation:

First of all split the fraction up like so:

`(x^2+2x-1)/(x^2+9) =x^2/(x^2+9)+(2x)/(x^2+9) -1/(x^2+9)`

We can now integrate each fraction one by one, i.e:

`int(x^2+2x-1)/(x^2+9)dx =intx^2/(x^2+9)+(2x)/(x^2+9) -1/(x^2+9)dx`

We will have to do a bit of rearranging of the first fraction (add and subtract 9):

`intx^2/(x^2+9)+(2x)/(x^2+9) -1/(x^2+9)dx=`

`=int(x^2+9-9)/(x^2+9)+(2x)/(x^2+9) -1/(x^2+9)dx`

This can now be rearranged like so:

`=int(x^2+9)/(x^2+9)+(2x)/(x^2+9) -1/(x^2+9)-9/(x^2+9)dx`

`=int1+(2x)/(x^2+9) -10/(x^2+9)dx`

-The first term obviously just integrates to `x`.

-For the second term we should apply: `int(f'(x))/f(x)=ln(f(x))+C`

`int (2x)/(x^2+9)dx=ln(x^2+9)+C`

-And for the third term:

`int10/(x^2+9)dx`

Use the substitution `x=3tan(u)`
`-> dx = 3 sec^2(u)du`

We also need the trig identity: `tan^2(x)+1=sec^2(x)` Putting the substitution in:

`10int(3sec^2(u))/(9tan^2(u)+9)du=10/3intsec^2(u)/sec^2(u)du`

`=10/3intdu=10/3u+C`

Reverse the substitution and we get:

`10/3tan^-1(x/3)+C`

So returning to our original integral if apply what have found we get that:

`int(x^2+2x-1)/(x^2+9)dx=int1+(2x)/(x^2+9) -10/(x^2+9)dx`

`=x+ln(x^2+9)-10/3tan^-1(x/3)+C`