How do you integrate (3x)/(x^2+1)?

2 Answers
Feb 19, 2015

The answer is: 3/2ln(x^2+1)+c.

Since:

int(f'(x))/(f(x))dx=ln|f(x)|+c

int(3x)/(x^2+1)dx=3/2int(2x)/(x^2+1)dx=3/2ln|x^2+1|+c=3/2ln(x^2+1)+c.

Feb 19, 2015

You can use a u-substitution to evaluate this integral

Let u=x^2+1 then du=2xdx therefore

(du)/2=xdx

Now substituting into the original integral we will have

int3/u(du)/2=3/2int(du)/u=3/2ln|u|

Now back substitute for u

3/2ln|x^2+1|+C