If $I_n=\int(\log x)^n d x$, then $I_n+n

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Indefinite Integration

Question:

If $I_n=\int(\log x)^n d x$, then $I_n+n I_{n-1}=$

Options:

$(x \log x)^n$

$x(\log x)^n$

$n(\log x)^n$

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

Correct Answer:

$x(\log x)^n$

Explanation:

Let $I_n=\int(\log x)^n . 1 d x$

$\Rightarrow I_n=x(\log x)^n-\int x n(\log x)^{n-1} \frac{1}{x} d x$

$\Rightarrow I_n=x(\log x)^n-n I_{n-1}$

$\Rightarrow I_n+n I_{n-1}=x(\log x)^n$