int {frac{{sqrt {{x^2} + 1} [log ({x^2} + 1) | vedclass

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $I = \int {\frac{{\sqrt {{x^2} + 1} [\log ({x^2} + 1) - 2\log x]}}{{{x^4}}}} dx$. We can rewrite the integrand as: $I = \int {\frac{{\sqrt {{x^2}(1 + 1/x^2)} [\log (x^2(1 + 1/x^2))]}}{{{x^4}}}} dx = \int {\frac{{x\sqrt {1 + 1/x^2} [\log x^2 + \log(1 + 1/x^2) - 2\log x]}}{{{x^4}}}} dx$. Since $\log x^2 = 2\log x$,the expression simplifies to: $I = \int {\sqrt {1 + \frac{1}{{{x^2}}}} \log \left( {1 + \frac{1}{{{x^2}}}} \right) \cdot \frac{1}{{{x^3}}}} dx$. Let $1 + \frac{1}{{{x^2}}} = t$. Then $-\frac{2}{{{x^3}}} dx = dt$,or $\frac{1}{{{x^3}}} dx = -\frac{1}{2} dt$. Substituting these into the integral: $I = - \frac{1}{2} \int {\sqrt t \log t} dt$. Using integration by parts,$\int u dv = uv - \int v du$,with $u = \log t$ and $dv = \sqrt t dt$: $I = - \frac{1}{2} \left[ \log t \cdot \frac{t^{3/2}}{3/2} - \int \frac{1}{t} \cdot \frac{t^{3/2}}{3/2} dt \right] = - \frac{1}{2} \left[ \frac{2}{3} t^{3/2} \log t - \frac{2}{3} \int t^{1/2} dt \right]$. $I = - \frac{1}{2} \left[ \frac{2}{3} t^{3/2} \log t - \frac{2}{3} \cdot \frac{t^{3/2}}{3/2} \right] + c = - \frac{1}{3} t^{3/2} \log t + \frac{2}{9} t^{3/2} + c$. $I = - \frac{1}{3} t^{3/2} \left[ \log t - \frac{2}{3} \right] + c$. Substituting $t = 1 + \frac{1}{{{x^2}}}$ back: $I = - \frac{1}{3} {\left( {1 + \frac{1}{{{x^2}}}} \right)^{3/2}} \left[ {\log \left( {1 + \frac{1}{{{x^2}}}} \right) - \frac{2}{3}} \right] + c$.

$\int {\frac{{\sqrt {{x^2} + 1} [\log ({x^2} + 1) - 2\log x]}}{{{x^4}}}} dx$ is equal to

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