Mathematics

# Solve : $\displaystyle \int \dfrac{1}{\sqrt{7 - 6x - x^2}} dx$

##### SOLUTION
$\displaystyle\int \dfrac{1}{\sqrt{7-6x-x^2}}dx$

$=\displaystyle\int \dfrac{1}{\sqrt{-(x+3)^2+16}}dx$

substitute $u=x+3\Rightarrow du=dx$

$=\displaystyle\int \dfrac{1}{\sqrt{-u^2+16}}du$

$=\sin^{-1}\dfrac{1}{4}(u)$

$\displaystyle\int \dfrac{1}{\sqrt{7-6x-x^2}}dx=\sin^{-1}\dfrac{1}{4}(x+3)+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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