1.Let I(x)=∫x2(xsec2x+tanx)(xtanx+1)2dxI(x) = \int \frac{x^2(x\sec^2 x+\tan x)}{(x\tan x+1)^2}dxI(x)=∫(xtanx+1)2x2(xsec2x+tanx)dx. If I(0)=0I(0)=0I(0)=0, then I(π/4)=I(\pi/4)=I(π/4)=a.\log_e\frac{(\pi+4)^2}{16} + \frac{\pi^2}{4(\pi+4)}b.\log_e\frac{(\pi+4)^2}{16} - \frac{\pi^2}{4(\pi+4)}c.\log_e\frac{(\pi+4)^2}{32} - \frac{\pi^2}{4(\pi+4)}d.\log_e\frac{(\pi+4)^2}{32} + \frac{\pi^2}{4(\pi+4)}Login to continueOnly logged in users canattempt or see the solution.
