How do you evaluate the integral of #int x/[(x^2)+1]dx#?

1 Answer
Jim H
Mar 22, 2016

Note that the derivative of the denominator is almost the numerator. Use substitution.

Explanation:

#int x/[(x^2)+1]dx#

Let #u=x^2+1#, so that #du = 2xdx#.

The integral is

#int x/[(x^2)+1]dx = int 1/(x^2+1) xdx#

# = 1/2 int underbrace(1/(x^2+1))_(1/u) underbrace(2xdx)_(du)#

# = 1/2 int 1/u du#

# = 1/2 ln abs u +C#

# = 1/2 ln abs(x^2+1)+C#

# = 1/2 ln(x^2+1)+C#

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