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−1ex−1dx=∫ex−1exdx−∫ex−11 ....(ii) Consider I1=∫ex−1exdx and I2=∫ex−11dx Now, I1=∫ex−1exdx Put ex−1=t ⇒ exdx=dt ∴ I1=∫tdt=2t+C1=2ex−1+C1 and I2=∫ex−11dx Put ex−1=z2 ⇒ exdx=2zdz⇒dx=z2+12zdz ∴ I2=∫z1⋅z2+12zdz=2∫z2+11dz=2tan−1z+C2=2tan−1ex−1+C2 ∵ I=I1−I2 [from Eq. (ii)] ∴ I=2ex−1+C1−2tan−1ex−1−C2=2ex−1−2tan−1ex−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=2 and B=−2 Hence, A+B=2+(−2)=2−2=0