If I=∫2−x2(1+x)1−x2dx then I is  equal to

Let I=∫2−x2(1+x)1−x2dxI=∫1−x2+1(1+x)1−x2dxI=∫1−x2(1+x)1−x2dx+∫dx(1+x)1−x2I=∫(1−x)dx1−x2dx+∫dx(1+x)1−x2I=∫11−x2dx+∫(−x)dx1−x2+∫dx(1+x)1−x2I=sin−1⁡x+1−x2+I1 Where I1=∫1(1+x)1−x2dx Put 1+x=1t⇒x=1t−1⇒dx=−1t2dtI1=∫−1t2dt1t1−1t−12=∫−1dtt2−1−2t+t2I1=∫−12t−1dt=−1222t−1+CI1=−21+x−1+C=−1−x1+x+C∴I=sin−1⁡x+1−x2−1−x1+x+C