I=∫(1+x)x1−xdxPut, 1+x=t⇒2x1dx=dt
And, 1−x=1−(t−1)2=2t−t2
∴I=∫t2t−t22dt
Again put, t=z1⇒dt=z2−1dz
⇒I=2∫z1z2−z21−z21dz=2∫2z−1−dz=−22z−1+c
Where c is arbitrary constant.
=−2t2−1+c
=−2t2−t+c
=−21+x1−x+c
The integral ∫(1+x)x−x2dx is equal to
Held on 10 Apr 2016 · Verified 6 Jul 2026.
−21−x1+x+c
−1+x1−x+c
−21+x1−x+c
1−x1+x+c
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