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

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Answer 1 · 2015-02-19T19:30:39

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#.

Answer 2 · 2015-02-19T19:31:32

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 #