Edexcel A level Maths (9MA0) · Pure mathematics › Integration
Practise Integration by parts. 4 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
A student is asked to find \(\int x\ln x\,\mathrm{d}x\). The student's working begins: \(u = x\), \(\frac{\mathrm{d}v}{\mathrm{d}x} = \ln x\) \(\Rightarrow\) \(\frac{\mathrm{d}u}{\mathrm{d}x} = 1\), \(v = \frac{1}{x}\) \(\int x\ln x\,\mathrm{d}x = x \times \frac{1}{x} - \int \frac{1}{x}\,\mathrm{d}x = 1 - \ln x + c\)
(a) Identify the mistake in the student's working.[1]
(b) Explain why \(u = \ln x\) and \(\frac{\mathrm{d}v}{\mathrm{d}x} = x\) is a better choice.[1]
(c) Find \(\int x\ln x\,\mathrm{d}x\).[4]
Show the answer and mark scheme
(a)Answer: \(v\) was found by differentiating \(\ln x\) instead of integrating it.
B1 for stating that \(v\) should be found by integrating \(\ln x\), not by differentiating it (\(\frac{1}{x}\) is the derivative of \(\ln x\))
Worked solution: If \(\frac{\mathrm{d}v}{\mathrm{d}x} = \ln x\) then \(v = \int \ln x\,\mathrm{d}x\), which is not \(\frac{1}{x}\); \(\frac{1}{x}\) is \(\frac{\mathrm{d}}{\mathrm{d}x}(\ln x)\).
(b)Answer: \(\ln x\) differentiates to \(\frac{1}{x}\), which cancels with the \(x^{2}\) from integrating \(x\), leaving an easy integral.
B1 for explaining that \(\ln x\) is easy to differentiate but not a standard integral, whereas \(x\) is easy to integrate, so the new integral \(\int \frac{x^{2}}{2} \times \frac{1}{x}\,\mathrm{d}x\) is simple
Worked solution: With \(u = \ln x\), \(\frac{\mathrm{d}u}{\mathrm{d}x} = \frac{1}{x}\) and \(v = \frac{1}{2}x^{2}\), so the remaining integral is \(\int \frac{1}{2}x\,\mathrm{d}x\), which is straightforward.
(c)Answer: \(\frac{1}{2}x^{2}\ln x - \frac{1}{4}x^{2} + c\)
M1 for integrating by parts with \(u = \ln x\), \(\frac{\mathrm{d}v}{\mathrm{d}x} = x\)
A1 for \(\frac{1}{2}x^{2}\ln x - \int \frac{1}{2}x\,\mathrm{d}x\)
dM1 for integrating the second term
A1 for \(\frac{1}{2}x^{2}\ln x - \frac{1}{4}x^{2} + c\)
Worked solution: \(\int x\ln x\,\mathrm{d}x = \frac{1}{2}x^{2}\ln x - \int \frac{1}{2}x^{2} \times \frac{1}{x}\,\mathrm{d}x = \frac{1}{2}x^{2}\ln x - \int \frac{1}{2}x\,\mathrm{d}x = \frac{1}{2}x^{2}\ln x - \frac{1}{4}x^{2} + c\)
M1 for integrating \(\mathrm{e}^{x}\sin x\) by parts with \(u = \sin x\): \(J = \left[\mathrm{e}^{x}\sin x\right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} \mathrm{e}^{x}\cos x\,\mathrm{d}x\)
A1 for \(\left[\mathrm{e}^{x}\sin x\right]_{0}^{\frac{\pi}{2}} = \mathrm{e}^{\frac{\pi}{2}}\)
A1* for \(J = \mathrm{e}^{\frac{\pi}{2}} - I\), so \(I + J = \mathrm{e}^{\frac{\pi}{2}}\) (cso)