\int_1^2 \frac{[loge(ex)]^n}{x}dx= where n≠ -1

Correct Answer is : [log2e]n+1n+1−logen+1

{[log2e]^{n+1}}/{n+1}- \frac{loge}{n+1}

CA CPT 2013 June Question Paper Quantitative Aptitude for online practice

Question : 24 of 50

Marks: +1, -0

∫_1^2 {[loge(ex)]^n}/{x}dx=

where n≠ -1

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