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

Question

Scarlett Allison · Accepted Answer

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

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

Charles Arocha · Answer

undefined To find the indefinite integral of $ \int \frac{x}{x^2 + 1} $, you can use the substitution method. Let $ u = x^2 + 1 $, then $ du = 2x \, dx $. Substituting these into the integral gives $ \frac{1}{2} \int \frac{du}{u} $. Integrating $ \frac{1}{u} $ gives $ \ln|u| $. Substituting back $ u = x^2 + 1 $ gives the final result $ \frac{1}{2} \ln|x^2 + 1| + C $, where $ C $ is the constant of integration.