Answer
$\displaystyle \frac{1}{2}\ln(\frac{18}{13})-\frac{π}{6}+\frac{2}{3}\tan^{-1}(\frac{2}{3})≈0.031114$
Work Step by Step
$\displaystyle \int_0^1\frac{x}{x^2+4x+13}dx$
Partial fraction decomposition cannot be done, therefore split the integral into two parts:
$\displaystyle \int_0^1\frac{x+2}{x^2+4x+13}-\frac{2}{x^2+4x+13}dx=\int_0^1\frac{x+2}{x^2+4x+13}-\int_0^1\frac{2}{x^2+4x+13}dx$
Use u-sub to evaluate the first integral:
$\displaystyle \int\frac{x+2}{x^2+4x+13}dx\quad$Let $\quad u=x^2+4x+13\quad\rightarrow\quad du=(2x+4)dx=2(x+2)dx$
$\displaystyle \frac{1}{2}\int\frac{du}{u}=\frac{1}{2}\ln|x^2+4x+13|+C$
Use inverse trigonometric substitution for the second integral
Recall that $\displaystyle \int\frac{du}{a^2+u^2}=\frac{1}{a}\tan^{-1}(\frac{u}{a})+C$
$\displaystyle \int\frac{2}{x^2+4x+13}dx=\int\frac{2}{(x^2+4x+4)+9}dx=\int\frac{2}{(x+2)^2+9}dx$
where $a^2=9$ and $u^2=(x+2)^2\quad\therefore\quad a=3$ and $u=x+2$
$\displaystyle 2[\frac{1}{3}\tan^{-1}(\frac{x+2}{3})]+C$
Now combine the two solutions together and evaluate them from $x=0$ to $x=1$
$\displaystyle [\frac{1}{2}\ln|x^2+4x+13|-\frac{2}{3}\tan^{-1}(\frac{x+2}{3})]_0^1$
$\displaystyle [\frac{1}{2}\ln|1^2+4(1)+13|-\frac{2}{3}\tan^{-1}(\frac{1+2}{3})] - [\frac{1}{2}\ln|0^2+4(0)+13|-\frac{2}{3}\tan^{-1}(\frac{0+2}{3})]$
$\displaystyle \frac{1}{2}\ln(18)-\frac{2}{3}\tan^{-1}(1)-\frac{1}{2}\ln(13)+\frac{2}{3}\tan^{-1}(\frac{2}{3})$
$\displaystyle \frac{1}{2}\ln(\frac{18}{13})-\frac{2}{3}(\frac{π}{4})+\frac{2}{3}\tan^{-1}(\frac{2}{3})$
$\displaystyle \frac{1}{2}\ln(\frac{18}{13})-\frac{π}{6}+\frac{2}{3}\tan^{-1}(\frac{2}{3})≈0.031114$
