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

1 Answer
Dec 4, 2016

# 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