Evaluate the integral $\int \frac{6+4x}{1+x^2}dx$

Question

Evaluate the integral $$\int \frac{6+4x}{1+x^2}\mathrm{d}x.$$

Hey guys, i've been trying to solve this integral and I'm pretty sure I'm supposed to use u-substitution only, but I can't get it to work out. So far I've tried setting $u=1+x^2$, $u=6+4x$, and other things but I can't get the answer (which involves ln and arctan).

I see where both $\arctan$ and $\ln$ could show up, as $$ \int \frac{1}{1+x^2}\mathrm{d}x = \arctan(x)+C.$$

Help appreciated!!

Answer 1

$$\frac{6+4x}{1+x^2} = \frac{6}{1+x^2} + \frac{4x}{1+x^2}$$

You should be able to integrate both parts.

Answer 2

Notice that you can split the integral to two integrals, in such a way that:

$$\int\frac{6+4x}{1+x^2}dx=\int\frac{6}{1+x^2}dx+\int\frac{4x}{1+x^2}dx=6\int\frac{1}{1+x^2}dx+\int\frac{4x}{1+x^2}dx=6\arctan(x)+\int\frac{4x}{1+x^2}dx$$

Now we define $u=1+x^2$, meaning $\frac{du}{dx}=2x$. Therefore we get: $$\int\frac{4x}{1+x^2}dx=\int\frac{4x}{u}\frac{1}{2x}du=\int\frac{2}{u}du=2\ln|u|=2\ln|1+x^2|.$$

And finally - $$\int\frac{6+4x}{1+x^2}dx=2\ln|1+x^2|+6\arctan(x)$$