# How do I evaluate int(3x)/(x^2+1)dx?

Mar 5, 2015

Use substitution (often called $u$ substitution).

Notice that the derivative of the denominator is $2 x$, which is a lot like what's in the numerator. The $3$ in the numerator is kinda in our way, so move it outside the integral sign.

(Recall that $\int c f \left(x\right) \mathrm{dx} = c \int f \left(x\right) \mathrm{dx}$, so this wont't change the integral.)

Now the integral is $3 \int \frac{x}{{x}^{2} + 1} \mathrm{dx}$.

Let $u = {x}^{2} + 1$ making $\mathrm{du} = 2 x \mathrm{dx}$and proceed with whatever details of substitution you've learned.

I use: $3 \int \frac{x}{{x}^{2} + 1} \mathrm{dx} = 3 \int \frac{1}{2} \frac{2 x}{{x}^{2} + 1} \mathrm{dx}$

$= \frac{3}{2} \int \frac{1}{{x}^{2} + 1} 2 x \mathrm{dx} = \frac{3}{2} \int \frac{1}{u} \mathrm{du}$

You can probably see how to finish to get $\frac{3}{2} \ln \left({x}^{2} + 1\right) + C$.