If I_n = int x^n cdot e^{cx} , dx for n geq 1 | vedclass

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(A) Given that,$I_n = \int x^n \cdot e^{cx} \, dx$. Using integration by parts,let $u = x^n$ and $dv = e^{cx} \, dx$. Then $du = n x^{n-1} \, dx$ and

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$x^n e^{cx}$

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(A) Given that,$I_n = \int x^n \cdot e^{cx} \, dx$. Using integration by parts,let $u = x^n$ and $dv = e^{cx} \, dx$. Then $du = n x^{n-1} \, dx$ and $v = \frac{e^{cx}}{c}$. Applying the formula $\int u \, dv = uv - \int v \, du$: $I_n = x^n \cdot \frac{e^{cx}}{c} - \int \frac{e^{cx}}{c} \cdot n x^{n-1} \, dx$. $I_n = \frac{x^n e^{cx}}{c} - \frac{n}{c} \int x^{n-1} e^{cx} \, dx$. Since $I_{n-1} = \int x^{n-1} e^{cx} \, dx$,we have: $I_n = \frac{x^n e^{cx}}{c} - \frac{n}{c} I_{n-1}$. Multiplying both sides by $c$: $c I_n = x^n e^{cx} - n I_{n-1}$. Rearranging the terms: $c I_n + n I_{n-1} = x^n e^{cx}$.

If $I_n = \int x^n \cdot e^{cx} \, dx$ for $n \geq 1$,then $c \cdot I_n + n \cdot I_{n-1}$ is equal to

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