If In=∫xn⋅eaxdx for n∈N,n≥1, then a⋅In+n⋅In−1=(where C is integration constant)

If In=∫xn⋅eaxdx for n∈N,n≥1, then a⋅In+n⋅In−1=(where C i

1.	If In=∫xn⋅eaxdx for n∈N,n≥1, then a⋅In+n⋅In−1=(where C is integration constant)
Answer» If $I_{n} = \int x^{n} \cdot e^{a x} d x$ for $n \in N, n \geq 1,$ then $a \cdot I_{n} + n \cdot I_{n - 1} =$ (where $C$ is integration constant)