Setting up this integral?

Question

$$\int\frac{2x+1}{9+x^2}dx$$

I tried to factor out the 9 to get $9(1+\frac{x^2}{9})=9(1+(\frac{x}{3})^2)$ to set up a u-sub to get arctan(x)..... But, it doesn't fit.

Is this integration by parts?

$u=9+x^2$

$du=2xdx$

$\frac{dv}{du}=2x+1$

$v=x^2dx$

What next?

Answer 1

Split your integral in two parts: $$\int \frac{2x+1}{9+x^2} dx = \int \frac{2x}{9+x^2} dx + \int \frac{1}{9+x^2} dx$$

For the first one, your $u$-substitution ($u=9+x^2$) will do the trick; you'll get a logarithm.

For the second one, work towards $\arctan(\ldots)$ just the way you started (factoring out 9).

Answer 2

$$\int\frac{2x+1}{9+x^2}dx\\=\int\frac{2x}{9+x^2}dx+\int \frac 1{9+x^2}dx\\=\ln|9+x^2|+\frac 13\arctan \frac x3+c$$

Answer 3

To the best of my knowledge trigonometric substitution would help if the numerator is a constant and when the denominator is a quadratic polynomial.

We retain our discussion on the case when the numerator is a linear polynomial and when the denominator is a quadratic polynomial. In an easier form $$\int \frac{c(x-a)}{(x-a)^2+b} dx,$$ we can just use the substitution $u = (x-a)^2 \implies du = 2(x-a)dx$ and then it becomes $$\int \frac{c}{2(u+b)} du = \frac{c}{2} \log \vert u + b \vert.$$ Putting back $u = (x-a)^2$ yields the result. In a more general form $$\int \frac{c(x-a)+d}{(x-a)^2+b} dx,$$ we split it into two integrals $$\int \frac{c(x-a)}{(x-a)^2+b} dx + \int \frac{d}{(x-a)^2+b} dx,$$ while we just discussed how to compute the first one and we just trigonometric substitution for the second one as we discussed in the first paragraph.