Integrate 1 x^2-6 x+13 with respect to x. - Vidyadip

Question

Integrate $\frac{1}{x^2-6 x+13}$ with respect to $x$.

✓

Answer

Here $
\begin{aligned}
x^2-6 x+13 & =x^2-6 x+3^2-3^2+13 \\
& =(x-3)^2+4
\end{aligned}
$
So, $\int \frac{d x}{x^2-6 x+13}=\int \frac{d x}{(x-3)^2+2^2}$
Let $x-3=t$ then $d x=d t$
$
\begin{aligned}
\text { So, } \int \frac{d x}{x^2-6 x+13}=\int & \frac{d t}{t^2+(2)^2} \\
=\frac{1}{2} \tan ^{-1}\left(\frac{t}{2}\right)+C \\
=\frac{1}{2} \tan ^{-1}\left(\frac{x-3}{2}\right)+C \text { }
\end{aligned}
$

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