Question

Question #ff170

Answer 1

`3ln(x^2+2x+2)-12tan^-1(x+1)+C`

Explanation:

`int(6x-6)/(x^2+2x+2)dx`

`=int(6x-6)/((x^2+2x+1)+1)dx`

`=int(6x-6)/((x+1)^2+1)dx`

In order to make the denominator resemble the trig identity `tan^2theta+1=sec^2theta`, let `tantheta=x+1`.

This also imply that `sec^2thetacolor(white).d theta=dx` and `x=tantheta-1`. Substituting these in gives:

`=int(6(tantheta-1)-6)/(tan^2theta+1)(sec^2thetacolor(white).d theta)`

`=int(6tantheta-12)/sec^2theta(sec^2thetacolor(white).d theta)`

`=int(6tantheta-12)d theta`

Both of these are standard integrals. If you forget the integration of `tantheta`, recall that `inttanthetacolor(white).d theta=intsintheta/costhetad theta`, then use the substitution `u=costheta`.

`=-6lnabscostheta-12theta+C`

Our original was substitution was `tantheta=x+1`, so `theta=tan^-1(x+1)`.

Furthermore, if `tantheta=x+1`, this is a triangle where the side opposite `theta` is `x+1` and the side adjacent is `1`, so the hypotenuse is `sqrt((x+1)^2+1)=sqrt(x^2+2x+2)`.

Then, `costheta="adjacent"/"hypotenuse"=1/sqrt(x^2+2x+2)`. So:

`=-6lnabs(1/sqrt(x^2+2x+2))-12tan^-1(x+1)`

Using `1/sqrt(x^2+2x+2)=(x^2+2x+2)^(-1/2)` and the log rule `ln(a^b)=bln(a)`, this becomes:

`= color(blue)(3ln(x^2+2x+2)-12tan^-1(x+1)+C`

The absolute value bars aren't necessary because `x^2+2x+2>0` for all real values of `x`.