Question

How do you find the indefinite integral of `int x/(x^2+1)`?

Answer 1

`int x/(x^2+1) dx = lnsqrt(x^2+1) + C`

Explanation:

With a little experience you may be able to see that the numerator is almost the derivative of the denominator, and we can use that:

`int x/(x^2+1) dx = 1/2int (2x)/(x^2+1) dx`
`:. int x/(x^2+1) dx = 1/2ln|x^2+1| + C`
`:. int x/(x^2+1) dx = 1/2ln(x^2+1) + C` (As `x^2+1>0`)
`:. int x/(x^2+1) dx = lnsqrt(x^2+1) + C`

If you can't spot that feature then we can use a substitution:

Let `u=x^2+1`, Then `(du)/dx = 2x`
So, "separating the variables" we get :

`int ... du = int ... 2xdx => int ... xdx = 1/2int ... du =`

And so substituting into the original integral:

`int x/(x^2+1) dx = 1/2int 1/u du`
`:. int x/(x^2+1) dx = 1/2ln|u| + C`
`:. int x/(x^2+1) dx = 1/2ln|x^2+1| + C`, as before