∫x2xsec2⁡x+tan⁡x(xtan⁡x+1)2dx is equal to

Let I=∫x2xsec2⁡x+tan⁡x(xtan⁡x+1)2dxUsing integrating by partsI=x2∫xsec2x+tanxxtanx+12dx−∫ddx(x2)∫xsec2x+tanxxtanx+12dxxtan⁡x+1=txsec2⁡x+tan⁡xdx=dtI=x2∫dtt2−∫2x∫dtt2dx=−x2t−∫2x−1tdx=−x2xtan⁡x+1+∫2x(xtan⁡x+1)dx=−x2xtan⁡x+1+∫2xcos⁡xxsin⁡x+cos⁡xdx Put xsin⁡x+cos⁡x=y(xcos⁡x+1sin⁡x−sin⁡x)=−x2xtan⁡x+1+2∫dtt=−x2xtan⁡x+1+2log⁡|xsin⁡x+cos⁡x|+C