The value of the integral int frac{sqrt{x^2+1 | vedclass

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Question

(A) Let $I = \int \frac{\sqrt{x^2+1} [\log(x^2+1) - 2 \log x]}{x^4} dx = \int \frac{x \sqrt{1+1/x^2} \log((1+1/x^2))}{x^4} dx = \int \frac{\sqrt{1+1/x

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$(1+\frac{1}{x^2})^{3/2} [-\frac{1}{3} \log(1+\frac{1}{x^2}) + \frac{2}{9}] + c$

Vedclass · Answer

(A) Let $I = \int \frac{\sqrt{x^2+1} [\log(x^2+1) - 2 \log x]}{x^4} dx = \int \frac{x \sqrt{1+1/x^2} \log((1+1/x^2))}{x^4} dx = \int \frac{\sqrt{1+1/x^2} \log(1+1/x^2)}{x^3} dx$.Let $t = 1+1/x^2$. Then $dt = -2/x^3 dx$, which implies $dx/x^3 = -dt/2$.Substituting these into the integral: $I = \int \sqrt{t} \log(t) (-1/2) dt = -1/2 \int t^{1/2} \log(t) dt$.Using integration by parts: $\int u dv = uv - \int v du$, where $u = \log(t)$ and $dv = t^{1/2} dt$.$du = (1/t) dt$ and $v = (2/3) t^{3/2}$.$I = -1/2 [\log(t) \cdot (2/3) t^{3/2} - \int (2/3) t^{3/2} \cdot (1/t) dt] = -1/2 [\frac{2}{3} t^{3/2} \log(t) - \frac{2}{3} \int t^{1/2} dt]$.$I = -1/2 [\frac{2}{3} t^{3/2} \log(t) - \frac{2}{3} \cdot \frac{2}{3} t^{3/2}] + c = -\frac{1}{3} t^{3/2} \log(t) + \frac{2}{9} t^{3/2} + c$.$I = t^{3/2} [-\frac{1}{3} \log(t) + \frac{2}{9}] + c = (1+\frac{1}{x^2})^{3/2} [-\frac{1}{3} \log(1+\frac{1}{x^2}) + \frac{2}{9}] + c$.

The value of the integral $\int \frac{\sqrt{x^2+1} [\log(x^2+1) - 2 \log x]}{x^4} dx$ is equal to...

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