फलन का समाकलन कीजिए: $\frac{\sqrt{x^{2}+1}\left[\log \left(x^{2}+1\right)-2 \log x\right]}{x^{4}}$

माना $I = \int \frac{\sqrt{x^{2}+1}\left[\log \left(x^{2}+1\right)-2 \log x\right]}{x^{4}} dx$
$\frac{\sqrt{x^{2}+1}\left[\log \left(x^{2}+1\right)-\log x^{2}\right]}{x^{4}} = \frac{\sqrt{x^{2}+1}}{x^{4}} \log \left(\frac{x^{2}+1}{x^{2}}\right) = \frac{\sqrt{x^{2}+1}}{x^{4}} \log \left(1+\frac{1}{x^{2}}\right)$
$= \frac{1}{x^{3}} \sqrt{\frac{x^{2}+1}{x^{2}}} \log \left(1+\frac{1}{x^{2}}\right) = \frac{1}{x^{3}} \sqrt{1+\frac{1}{x^{2}}} \log \left(1+\frac{1}{x^{2}}\right)$
माना $1+\frac{1}{x^{2}} = t$. तब $-\frac{2}{x^{3}} dx = dt$,अतः $\frac{1}{x^{3}} dx = -\frac{1}{2} dt$.
$I = -\frac{1}{2} \int \sqrt{t} \log t \, dt = -\frac{1}{2} \int t^{\frac{1}{2}} \log t \, dt$
खंडशः समाकलन (Integration by parts) का उपयोग करने पर: $\int u v \, dt = u \int v \, dt - \int (u' \int v \, dt) dt$
$I = -\frac{1}{2} \left[ \log t \cdot \frac{t^{3/2}}{3/2} - \int \frac{1}{t} \cdot \frac{t^{3/2}}{3/2} \, dt \right] = -\frac{1}{2} \left[ \frac{2}{3} t^{3/2} \log t - \frac{2}{3} \int t^{1/2} \, dt \right]$
$I = -\frac{1}{3} t^{3/2} \log t + \frac{1}{3} \cdot \frac{2}{3} t^{3/2} + C = -\frac{1}{3} t^{3/2} \log t + \frac{2}{9} t^{3/2} + C$
$t = 1+\frac{1}{x^{2}}$ प्रतिस्थापित करने पर:
$I = -\frac{1}{3} \left(1+\frac{1}{x^{2}}\right)^{3/2} \left[ \log \left(1+\frac{1}{x^{2}}\right) - \frac{2}{3} \right] + C$