∫dx(1+x)x−x2 is equal to

Let I=∫dx(1+x)x−x2 If x=sin⁡p, then 12xdx=cos⁡pdpI=∫2sin⁡pcos⁡pdp(1+sin⁡p)sin⁡p⋅cos⁡p=2∫dp(1+sin⁡p)=2∫(1−sin⁡p)dpcos2⁡p=2∫sec2⁡pdp−∫(tan⁡psec⁡p)dp=2(tan⁡p−sec⁡p)=2x(1−x)−1(1−x))+C=2(x−1)(1−x)+C