Evaluate the following integrals: 1 x^2+6x+13 dx - Vidyadip

Question

Evaluate the following integrals:
$\int\frac{1}{\text{x}^2+6\text{x}+13}\text{dx}$

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Answer

We have $\text{x}^2+6\text{x}+13=\text{x}^2+6\text{x}+3^2-3^2+13=(\text{x}+3)^2+4$
Sol, $\int\frac{\text{dx}}{\text{x}^2+6\text{x}+13}=\int\frac{1}{(\text{x}+3)^2+2^2}\text{dx}$
Let $\text{x}+3=\text{t}$ Then $\text{dx = dt}$
Therefore, $\int\frac{\text{dx}}{\text{x}^2+6\text{x}+13}=\int\frac{\text{dt}}{\text{t}^2+2^2}=\frac{1}{2}\tan^{-1}\frac{\text{t}}{2}+\text{c}$ [by 7.4 (3)]
$=\frac{1}{2}\tan^{-1}\frac{\text{x}+3}{2}+\text{C}$

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