First slide

Methods of integration

Question

$If I_{n} = \int (\log x)^{n} dx, then I_{n} + {nI}_{n - 1}$ is equal to

Moderate

A

$x (\log x)^{n}$
B

$(xlog x)^{n}$
C

$(\log x)^{n - 1}$
D

$n (\log x)^{n}$

Solution

$\begin{array}{l} I_{n} = \int (\log x)^{n} dx \\ ∴ I_{n - 1} = \int (\log x)^{n - 1} dx \\ Now, \begin{aligned} I_{n} & = \int (\log x)^{n} \cdot 1 dx \\ = (\log x)^{n} x - n \int (\log x)^{n - 1} \frac{1}{x} xdx \\ = x (\log x)^{n} - n \int (\log x)^{n - 1} dx \\ \Rightarrow I_{n} & = x (\log x)^{n} - {nI}_{n - 1} \\ ∴ I_{n} + {nI}_{n - 1} & = x (\log x)^{n} \end{aligned} \end{array}$

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