Evaluate the following integrals: 1 16-6x-x^2 dx - Vidyadip

Question

Evaluate the following integrals:$\int\frac{1}{\sqrt{16-6\text{x}-\text{x}^2}}\text{ dx}$

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Answer

$\int\frac{\text{dx}}{\sqrt{16-6\text{x}-\text{x}^2}}$
$=\int\frac{\text{dx}}{\sqrt{16-(\text{x}^2+6\text{x})}}$
$=\int\frac{\text{dx}}{\sqrt{16-(\text{x}^2+6\text{x}+3^2-3^2)}}$
$=\int\frac{\text{dx}}{\sqrt{16+9-(\text{x}+3)^2}}$
$=\int\frac{\text{dx}}{\sqrt{5^2-(\text{x}+3)^2}}$
$=\sin^{-1}\Big(\frac{\text{x}+3}{5}\Big)+\text{C}$

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