If f(x)=x,g(x)=ex−1 and h(x)=tan−1x
then the anti derivative of (f o g) (x) is
2[(g∘f)(x)−(g∘f∘h)(x)]+C
2[(f∘g)(x)−(f∘g∘h)(x)]+C
(f∘g)(x)+(f∘g∘h)(x)+C
2[(f∘g)(x)−(h∘f∘g)(x)]+C
I=∫ex−1dx
Put ex−1=t2, so that exdx=2tdt
∴I=∫t2tt2+1dt=2∫t2+1−1t2+1dt=2t−tan−1t+C=2ex−1−tan−1ex−1+C
=2[( fog )(x)−( hofog )(x)]+C.