Question
Evaluate the following integrals:
$\int\frac{1}{\text{x}^2-10\text{x}+34}\text{dx}$

Answer

Let $\text{I}=\int\frac{1}{\text{x}^2-10\text{x}+34}\text{dx}$
$=\int\frac{1}{\text{x}^2-2\text{x}\times5+(5)^2-(5)^2+34}\text{dx}$
$=\int\frac{1}{(\text{x}-5)^2+9}\text{dx}$
Let $(\text{x}-1)=\text{t} \dots(1)$
$\Rightarrow\text{dx = dt}$
so,
$\text{I}=\int\frac{1}{\text{t}^2+(3)^2}\text{dt}$
$\text{I}=\frac{1}{3}\tan^{-1}\big(\frac{\text{t}}{3}\big)+\text{C}$ $\Big[\text{since,}\int\frac{1}{\text{x}^2+\text{a}^2}\text{dx}=\frac{1}{\text{a}}\tan^{-1}\big(\frac{\text{x}}{2}\big)+\text{C}\Big]$
$\text{I}=\frac{1}{3}\tan^{-1}\Big(\frac{\text{x}-5}{3}\Big)+\text{C}$ [using (1)]

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