If $I_{m, n}=\int e^{m x} \cdot x^n d x$, then $I_{m, n}+\frac{n}{m} I_{m, n-1}=$

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Aicharya · Accepted Answer

$\begin{aligned} & \because I_{m, n}=\int e^{m x} \cdot x^n d x \ & =\frac{1}{m} x^n e^{m x}-\int\left(n x^{n-1}\right)\left(\frac{e^{m x}}{m}\right) d x \ & =\frac{x^n e^{m x}}{m}-\frac{n}{m} \int e^{m x} x^{n-1} d x=\frac{x^n e^{m x}}{m}-\frac{n}{m} I_{m, n-1}+c \ & \Rightarrow I_{m, n}+\frac{n}{m} I_{m, n-1}=\frac{x^n e^{m x}}{m}+c \end{aligned}$ Hence, option (c) is correct.

If \(I_{m, n}=\int e^{m x} \cdot x^n d x\), then \(I_{m, n}+\frac{n}{m} I_{m, n-1}=\)

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