int frac{x cdot log x}{left(sqrt{x^2-1}right) | vedclass

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

Question

(B) Let $I = \int \frac{x \log x}{(x^2-1)^{3/2}} dx$. Using integration by parts,let $u = \log x$ and $dv = \frac{x}{(x^2-1)^{3/2}} dx$. Then $du = \f

Vedclass · Accepted Answer

$\sec ^{-1} x-\frac{\log x}{\sqrt{x^2-1}}+C$

Vedclass · Answer

(B) Let $I = \int \frac{x \log x}{(x^2-1)^{3/2}} dx$. Using integration by parts,let $u = \log x$ and $dv = \frac{x}{(x^2-1)^{3/2}} dx$. Then $du = \frac{1}{x} dx$ and $v = \int (x^2-1)^{-3/2} x dx = -(x^2-1)^{-1/2} = -\frac{1}{\sqrt{x^2-1}}$. Using the formula $\int u dv = uv - \int v du$: $I = -\frac{\log x}{\sqrt{x^2-1}} - \int \left(-\frac{1}{\sqrt{x^2-1}}\right) \frac{1}{x} dx$ $I = -\frac{\log x}{\sqrt{x^2-1}} + \int \frac{1}{x \sqrt{x^2-1}} dx$ Since $\int \frac{1}{x \sqrt{x^2-1}} dx = \sec^{-1} x + C$,$I = \sec^{-1} x - \frac{\log x}{\sqrt{x^2-1}} + C$.

$\int \frac{x \cdot \log x}{\left(\sqrt{x^2-1}\right)^3} d x=$

Similar Questions

$\int (1 - x^2) \log x \, dx = $

$\int x \cdot \frac{\ln(x + \sqrt{1 + x^2})}{\sqrt{1 + x^2}} \, dx$ equals :

$\int \frac{x^3}{\sqrt{1+x^2}} dx$ is equal to

If $\int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x=f(x)-\log \left(1+x^2\right)$, then $f(x)$ is equal to

Find $\int e^{x} \sin x \, dx$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module