Answer
$$\int_{0}^{1} \frac{x}{x^{2}+4 x+13} d x=\frac{1}{2} \ln \frac{18}{13}-\frac{\pi}{6}+\frac{2}{3} \tan ^{-1} \frac{2}{3}$$
Work Step by Step
Given $$\int_{0}^{1} \frac{x}{x^{2}+4 x+13} d x$$
So, we have
\begin{aligned} I=& \int_{0}^{1} \frac{x}{x^{2}+4 x+13} d x \\=& \frac{1}{2} \int_{0}^{1} \frac{2 x}{x^{2}+4 x+13} d x \\=& \frac{1}{2} \int_{0}^{1} \frac{2 x+4}{x^{2}+4 x+13}-\frac{4}{x^{2}+4 x+13} d x \\=& \frac{1}{2} \int_{0}^{1} \frac{2 x+4}{x^{2}+4 x+13}-\frac{4}{(x+2)^{2}+9} d x \\=& \frac{1}{2} \int_{0}^{1} \frac{2 x+4}{x^{2}+4 x+13} d x-\frac{1}{2} \int_{0}^{1} \frac{4}{(x+2)^{2}+9} d x \\
&=\frac{1}{2} \left[\ln |x^2+4x+13|\right]_{0}^{1}-2\int_{0}^{1} \frac{1}{(x+2)^{2}+9} d x\end{aligned}
Since $$\int \frac{d x}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C$$
So, we get
\begin{align} &I= \frac{1}{2} \left[\ln |x^2+4x+13|\right]_{0}^{1}-2\int_{0}^{1} \frac{1}{(x+2)^{2}+3^2} d x\\
&= \frac{1}{2} \ln 18-\frac{1}{2} \ln 13-\left[\frac{2}{3} \tan ^{-1}\left[\frac{x+2}{3}\right]\right]_{0}^{1} \\
& =\frac{1}{2} \ln \frac{18}{13}-\frac{2}{3}\left[\tan ^{-1}1-\tan ^{-1} \frac{2}{3}\right] \\
& =\frac{1}{2} \ln \frac{18}{13}-\frac{2}{3}\left[\frac{\pi}{4}-\tan ^{-1} \frac{2}{3}\right] \\
& =\frac{1}{2} \ln \frac{18}{13}-\frac{\pi}{6}+\frac{2}{3} \tan ^{-1} \frac{2}{3} \end{align}