Evaluate the integral: int frac{sqrt{x^2+1}le | vedclass

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(A) Let $I = \int \frac{\sqrt{x^2+1}\left[\log \left(x^2+1\right)-2 \log x\right]}{x^4} d x$. We can rewrite the integrand as: $I = \int \frac{x\sqrt{

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

Vedclass · Answer

(A) Let $I = \int \frac{\sqrt{x^2+1}\left[\log \left(x^2+1\right)-2 \log x\right]}{x^4} d x$. We can rewrite the integrand as: $I = \int \frac{x\sqrt{1+\frac{1}{x^2}} \left[\log \left(x^2(1+\frac{1}{x^2})\right)-2 \log x\right]}{x^4} d x$ $I = \int \frac{\sqrt{1+\frac{1}{x^2}} \left[\log x^2 + \log(1+\frac{1}{x^2}) - 2 \log x\right]}{x^3} d x$ Since $\log x^2 = 2 \log x$,the expression simplifies to: $I = \int \frac{\sqrt{1+\frac{1}{x^2}} \log(1+\frac{1}{x^2})}{x^3} d x$. Let $1+\frac{1}{x^2} = t^2$. Then $-\frac{2}{x^3} dx = 2t dt$,which implies $-\frac{dx}{x^3} = t dt$. Substituting these into the integral: $I = -\int t \cdot \log(t^2) \cdot t dt = -\int t^2 \cdot 2 \log t dt = -2 \int t^2 \log t dt$. Using integration by parts: $I = -2 \left[ \frac{t^3}{3} \log t - \int \frac{t^3}{3} \cdot \frac{1}{t} dt \right] = -2 \left[ \frac{t^3}{3} \log t - \frac{t^3}{9} \right] = \frac{2}{9} t^3 - \frac{2}{3} t^3 \log t$. $I = \frac{1}{9} t^3 [2 - 6 \log t] = \frac{1}{9} t^3 [2 - 3 \log t^2]$. Substituting $t^2 = 1+\frac{1}{x^2}$ and $t = (1+\frac{1}{x^2})^{1/2}$: $I = \frac{1}{9} (1+\frac{1}{x^2})^{3/2} [2 - 3 \log (1+\frac{1}{x^2})] + c$.

Evaluate the integral: $\int \frac{\sqrt{x^2+1}\left[\log \left(x^2+1\right)-2 \log x\right]}{x^4} d x$

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