The value of $\int x \log x(\log x-1) d

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Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Indefinite Integration

Question:

The value of $\int x \log x(\log x-1) d x$ is equal to

Options:

$2(x \log x-x)^2+C$

$\frac{1}{2}(x \log x-x)^2+C$

$(x \log x)^2+C$

$\frac{1}{2}(x \log x)^3+C$

Correct Answer:

$\frac{1}{2}(x \log x-x)^2+C$

Explanation:

Let

$I=\int x \log x(\log x-1) d x=\int \log x(x \log x-x) d x$

$\Rightarrow I=\int(x \log x-x) d(x \log x-x)=\frac{(x \log x-x)^2}{2}+C$