I=∫2−x2+2−x2xdx Put t=2−x2,dxdt=22−x21⋅(−2x)⇒−tdt=xdx∴I=∫t2+t(−t)dt=−∫t+11dt=−log∣t+1∣=−log2−x2+1+c
JEE Main 2013 — Mathematics Calculus
The integral ∫2−x2+2−x2xdx equals :
Held on 23 Apr 2013 · Verified 6 Jul 2026.
log1+2+x2+c
−log1+2−x2+c
−xlog1−2−x2+c
xlog1−2+x2+c
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