Question

How do you integrate `int (x+4)/(x^2+2x+4)` using trigonometric substitution?

Answer 1

`sqrt3arctan((x+1)/sqrt3)+1/2ln(x^2+2x+4)+C`

Explanation:

We have:

`int(x+4)/(x^2+2x+4)dx=int(x+4)/((x+1)^2+3)dx`

Let `x+1=sqrt3tantheta`. Thus, `dx=sqrt3sec^2thetad theta`. Also note that this means `x+4=3+sqrt3tantheta`. This yields:

`=int(3+sqrt3tantheta)/(3tan^2theta+3)(sqrt3sec^2thetad theta)`

`=int(3sqrt3+3tantheta)/(3(tan^2theta+1))(sec^2thetad theta)`

Cancel the `3` terms and recall that tan^2theta+1=sec^2theta#:

`=int(sqrt3+tantheta)/sec^2theta(sec^2thetad theta)`

`=int(sqrt3+tantheta)d theta`

These are common integrals:

`=sqrt3theta+lnabssectheta+C`

Note that `theta=arctan((x+1)/sqrt3)`. Also, we can draw the right triangle where `tantheta=(x+1)/sqrt3` to see that `sectheta=sqrt((x+1)^2+3)/sqrt3=sqrt((x^2+2x+4)/3)`. Thus:

`=sqrt3arctan((x+1)/sqrt3)+lnabssqrt((x^2+2x+4)/3)+C`

Using the log rule `log(A^B)=Blog(A)`:

`=sqrt3arctan((x+1)/sqrt3)+1/2lnabs((x^2+2x+4)/3)+C`

Using the `log(A/B)=log(A)-log(B)` the `-1/2ln(3)` will come out of the natural logarithm and be absorbed into `C`:

`=sqrt3arctan((x+1)/sqrt3)+1/2lnabs(x^2+2x+4)+C`

Since `x^2+2x+4>0` for all (real) values of `x`:

`=sqrt3arctan((x+1)/sqrt3)+1/2ln(x^2+2x+4)+C`