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

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(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 $\int \frac{1}{x \sqrt{x^2-1}} dx = \sec^{-1} x + C$,$I = \sec^{-1} x - \frac{\log x}{\sqrt{x^2-1}} + C$.

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

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