int frac{2 x+2}{sqrt{x^2-4 x-5}} d x is equal | vedclass

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(A) Let $I = \int \frac{2x+2}{\sqrt{x^2-4x-5}} dx$. We express the numerator as $2x+2 = A \frac{d}{dx}(x^2-4x-5) + B$. $2x+2 = A(2x-4) + B$. Comparing

Vedclass · Accepted Answer

$2 \sqrt{x^2-4 x-5}+6 \log \left|(x-2)+\sqrt{x^2-4 x-5}\right|+C$

Vedclass · Answer

(A) Let $I = \int \frac{2x+2}{\sqrt{x^2-4x-5}} dx$. We express the numerator as $2x+2 = A \frac{d}{dx}(x^2-4x-5) + B$. $2x+2 = A(2x-4) + B$. Comparing coefficients of $x$ and constants,we get $2A = 2 \Rightarrow A = 1$ and $-4A + B = 2 \Rightarrow -4(1) + B = 2 \Rightarrow B = 6$. Thus,$I = \int \frac{2x-4}{\sqrt{x^2-4x-5}} dx + 6 \int \frac{dx}{\sqrt{(x-2)^2 - 3^2}}$. For the first integral,let $t = x^2-4x-5$,then $dt = (2x-4)dx$. $I = \int t^{-1/2} dt + 6 \int \frac{dx}{\sqrt{(x-2)^2 - 3^2}}$. $I = 2\sqrt{t} + 6 \log |(x-2) + \sqrt{(x-2)^2 - 3^2}| + C$. $I = 2\sqrt{x^2-4x-5} + 6 \log |(x-2) + \sqrt{x^2-4x-5}| + C$.

$\int \frac{2 x+2}{\sqrt{x^2-4 x-5}} d x$ is equal to

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