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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

$\int_{0}^{\infty} e^{-x} \sin^{6} x dx =$

If the graph of the anti-derivative $g(x)$ of $f(x) = \log(\log x) + (\log x)^{-2}$ passes through $(e, 2023 - e)$ and the term independent of $x$ in $g(x)$ is $k$,then the sum of all the digits of $k$ is

$\int \frac{d x}{(x+1) \sqrt{x^2+4}} = $

If $\int e^{\sin ^2 x}(\sin x \cos x+\cos ^3 x \sin x) d x = e^{\sin ^2 x}(1+f(x))+c$, then $f^{\prime}(x)=$

The value of $\int \frac{d x}{x^2\left(x^4+1\right)^{\frac{3}{4}}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module