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Paper 2 · November 2011 · Integration

Find $\int x^3\ln x\,dx$ .

Model answer

(x^4/4)ln x - x^4/16

Also accepted: x^4/4 ln x - x^4/16, (1/4)x^4 ln x - (1/16)x^4

Explanation

Integrate by parts with

v=\ln x

and

\frac{du}{dx}=x^3

, so

u=\frac{x^4}{4}

. Then

\int x^3\ln x\,dx=\frac{x^4}{4}\ln x-\int\frac{x^4}{4}\cdot\frac1x\,dx=\frac{x^4}{4}\ln x-\int\frac{x^3}{4}\,dx=\frac{x^4}{4}\ln x-\frac{x^4}{16}

.

Derived from ZIMSEC Mathematics Paper 2, November 2011, Q1

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