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

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Answer 1 · 2016-03-22T12:26:48

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

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

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

The integral is

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

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

$= 1/2 int 1/u du$

$= 1/2 ln abs u +C$

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

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

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