Integration of rational functions How would you integrate a rational function like this:
$$\frac{2x+3}{x^2+2x+10}$$
 A: HINT:
As $x^2+2x+10=(x+1)^2+3^2,$
write $2x+3=A(x+1)+B$
Now use $\dfrac{d(\arctan x)}{dx}=?$
A: Notice that:
$$\frac{2x + 3}{x^2 + 2x + 10} = \frac{2x + 2}{x^2 + 2x + 10} + \frac{ 1}{x^2 + 2x + 10} = \frac{2x + 2}{x^2 + 2x + 10} + \frac{ 1}{(x+1)^2 + 3^2}$$
Therefore, 
$$\int \frac{2x + 3}{x^2 + 2x + 10} dx = \int\frac{2x + 2}{x^2 + 2x + 10}dx+ \int\frac{ 1}{(x+1)^2 + 3^2}dx$$
$$ = \ln|x^2 + 2x + 10| +  \int\frac{ 1}{(x+1)^2 + 3^2}dx $$
$$= \ln|x^2 + 2x + 10| +   \int\frac{1/9}{(\frac{x+1}{3})^2 + 1}dx $$
$$= \ln|x^2 + 2x + 10| +   \int\frac{1/9}{(\frac{x+1}{3})^2 + 1}dx$$ 
$$= \ln|x^2 + 2x + 10| +   \frac{1}{9}\int\frac{1}{(\frac{x+1}{3})^2 + 1}dx$$ 
$$= \ln|x^2 + 2x + 10| +   \frac{1}{3} \arctan(\frac{x+1}{3}) + c$$ 
We can safely drop the absolute value, because $x^2 + 2x + 10 > 0 \quad \forall x \in \mathbb{R}$, obtaining:
$$\int \frac{2x + 3}{x^2 + 2x + 10} dx = \ln(x^2 + 2x + 10) +   \frac{1}{3} \arctan(\frac{x+1}{3}) + c$$ 
A: $x^{2}+2x+10$ prime $\rightarrow$
$$I=\int\dfrac{2x+3}{x^{2}+2x+10}dx=\int \dfrac{2x+2+1}{x^{2}+2x+10}dx=\int \dfrac{2x+2}{x^{2}+2x+10}dx +\int \dfrac{1}{(x^{2}+2x+1)+9}dx=\ln (x^{2}+2x+10)+J$$
$$J=\int \dfrac{1}{(x+1)^{2}+3^{2}}dx=\int \dfrac{1}{9\left[ \left ( \dfrac{x+1}{3}\right)^{2}+1\right]}dx=\dfrac{1}{3}\int \dfrac{\dfrac{1}{3}}{\left( \dfrac{x+1}{3}\right)^{2}+1}dx=\dfrac{1}{3}\arctan \left( \dfrac{x+1}{3}\right)$$
$$I=\ln (x^{2}+2x+10)+\dfrac{1}{3}\arctan \left( \dfrac{x+1}{3}\right)+C$$
A: $$\int \frac{2x+3}{x^2+2x+10} dx$$
$$= \int \frac{2x+2+1}{x^2+2x+10} dx $$
$= \int \frac{2x+2}{x^2+2x+10} dx + \int \frac{1}{x^2+2x+10}dx$
Let solve them individually,
$$\int \frac{2x+2}{x^2+2x+10} dx$$
$$\int \frac{f'(x)}{f(x)} dx = \log|f(x)|$$
$$= \log |x^2 + 2x + 10| + c$$
Other part -
$$\int \frac{1}{x^2+2x+10} dx$$
Use completing the square -
$$\int \frac{1}{x^2+2x+1+9} dx$$
$$=\int \frac{1}{(x+1)^2+(3)^2} dx$$
$$\int \frac 1{x^2+a^2} dx = \frac 1a tan^{-1}\frac xa + c$$
$$=\frac 13 tan^{-1}\frac {x+1}3+ c$$
Add both to get complete answer.
