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If $\mathrm{n} \in \mathrm{N}$ and $\mathrm{I}_{\mathrm{n}}=\int(\log \mathrm{x})^{\mathrm{n}} \mathrm{dx}$, then $\mathrm{I}_{\mathrm{n}}+\mathrm{n} \mathrm{} \mathrm{I}_{\mathrm{n}-1}$ is equal to
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The correct answer is:
$x(\log x)^{n}+C$
Here, $\mathrm{I}_{\mathrm{n}}=\int(\log \mathrm{x})^{\mathrm{n}} \mathrm{dx}$
On initegrating by parts, we get
$$
\begin{aligned}
&\mathrm{I}_{\mathrm{n}}=(\log \mathrm{x})^{\mathrm{n}} \cdot \mathrm{x}-\int \mathrm{x} \cdot \mathrm{n}(\log \mathrm{x})^{\mathrm{n}-1} \frac{1}{\mathrm{x}} \mathrm{dx} \\
&\mathrm{I}_{\mathrm{n}}=\mathrm{x}(\log \mathrm{x})^{\mathrm{n}} \mathrm{nI}_{\mathrm{n}-1} \\
&\therefore \quad \mathrm{I}_{\mathrm{n}}+\mathrm{nI}_{\mathrm{n}-1}=\mathrm{x}(\log \mathrm{x})^{\mathrm{n}}+\mathrm{C}
\end{aligned}
$$
On initegrating by parts, we get
$$
\begin{aligned}
&\mathrm{I}_{\mathrm{n}}=(\log \mathrm{x})^{\mathrm{n}} \cdot \mathrm{x}-\int \mathrm{x} \cdot \mathrm{n}(\log \mathrm{x})^{\mathrm{n}-1} \frac{1}{\mathrm{x}} \mathrm{dx} \\
&\mathrm{I}_{\mathrm{n}}=\mathrm{x}(\log \mathrm{x})^{\mathrm{n}} \mathrm{nI}_{\mathrm{n}-1} \\
&\therefore \quad \mathrm{I}_{\mathrm{n}}+\mathrm{nI}_{\mathrm{n}-1}=\mathrm{x}(\log \mathrm{x})^{\mathrm{n}}+\mathrm{C}
\end{aligned}
$$
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