∫13dxx2(x2+1)2=∫13(1x2−x2+2x4+2x2+1)dx=∫13(1x2−1x2+1−1(x2+1)2)dx\int_1^{\sqrt{3}} \frac{dx}{x^2(x^2+1)^2} = \int_1^{\sqrt{3}} \left(\frac{1}{x^2} - \frac{x^2+2}{x^4+2x^2+1}\right)dx = \int_1^{\sqrt{3}} \left(\frac{1}{x^2} - \frac{1}{x^2+1} - \frac{1}{(x^2+1)^2}\right)dx∫13x2(x2+1)2dx=∫13(x21−x4+2x2+1x2+2)dx=∫13(x21−x2+11−(x2+1)21)dx =[−1x−tan−1x−12(xx2+1+tan−1x)]13=−13−π3−12(34+π3)+1+π4+12(12+π4)= \left[-\frac{1}{x} - \tan^{-1}x - \frac{1}{2}\left(\frac{x}{x^2+1}+\tan^{-1}x\right)\right]_1^{\sqrt{3}} = -\frac{1}{\sqrt{3}} - \frac{\pi}{3} - \frac{1}{2}\left(\frac{\sqrt{3}}{4}+\frac{\pi}{3}\right) + 1 + \frac{\pi}{4} + \frac{1}{2}\left(\frac{1}{2}+\frac{\pi}{4}\right)=[−x1−tan−1x−21(x2+1x+tan−1x)]13=−31−3π−21(43+3π)+1+4π+21(21+4π) =54−11324−π8= \frac{5}{4} - \frac{11\sqrt{3}}{24} - \frac{\pi}{8}=45−24113−8π