We have, f(x)=√x and g(x)=ex−1 ∴ fog(x)=f{g(x)}=f{ex−1} ⇒ fog(x)=√ex−1 ....(i) Let I=∫fog(x)‌dx =∫√ex−1‌dx [from Eq. (i)] =∫
ex−1
√ex−1
‌dx =∫
ex
√ex−1
‌dx−∫
1
√ex−1
....(ii) Consider I1=∫
ex
√ex−1
‌dx and I2=∫
1
√ex−1
‌dx Now, I1=∫
ex
√ex−1
‌dx Put ex−1=t ⇒ ex‌dx=dt ∴ I1=∫
dt
√t
=2√t+C1=2√ex−1+C1 and I2=∫
1
√ex−1
‌dx Put ex−1=z2 ⇒ ex‌dx=2‌zdz⇒dx=
2z
z2+1
‌dz ∴ I2=∫
1
z
.
2z
z2+1
‌dz=2‌∫
1
z2+1
‌dz =2tab−1z+C2=2tan−1√ex−1+C2 ∵ I=I1−I2 [from Eq. (ii)] ∴ I=2√ex−1+C1−2tan−1√ex−1−C2 =2√ex−1−2tan−1√ex−1+C[where,C=C1−C2] =2‌fog(x)−2tan−1‌fog(x)+C[∵fog(x)=√ex−1] Now, comparing with the given integral, we get A=2‌and‌B=−2 Hence, A+B=2+(−2)=2−2=0