The correct option is D 12√p3+1ln∣∣
∣∣ex+x−√p3+1ex+x+√p3+1∣∣
∣∣+C for p>−1
I=∫(ex+1)dxe2x+x2+2xex−p3−1
=∫(ex+1)dx(ex+x)2−p3−1
Put ex+x=t⇒(ex+1)dx=dt
∴I=∫dtt2−p3−1
For p=−1,
I=∫dtt2
=−1t+C
=−1ex+x+C
For p<−1,
I=∫dtt2+(−p3−1)+ve
=1√−p3−1tan−1(ex+x√−p3−1)+C
For p>−1,
I=∫dtt2−(p3+1)+ve
=12√p3+1ln∣∣
∣∣ex+x−√p3+1ex+x+√p3+1∣∣
∣∣+C