2023TrialShort answer

Q12 (c) (i)

2 marks

Question

Let

I_n=\int_{e^{-1}}^1(1+\log_e x)^n\,dx

and

J_n=\int_{e^{-1}}^1(\log_e x)(1+\log_e x)^n\,dx

for $n=0,1,2,3,\ldots$ .

Show that $I_n=1-nI_{n-1}$ for $n=1,2,3,\ldots$ .

Try the question before revealing the answer.

Solution

(c)(i)

I_n=\int_{e^{-1}}^1(1+\log_e x)^n\,dx

u=(1+\ln x)^n,\qquad v'=1

u'=\frac{n}{x}(1+\ln x)^{n-1},\qquad v=x

\begin{aligned} I_n&=\left[x(1+\ln x)^n\right]_{e^{-1}}^1-\int_{e^{-1}}^1e^{-x}\left(\frac nx\right)(1+\ln x)^{n-1}\,dx\\ &=(1+\ln(1))^n-e^{-1}(1+\ln e^{-1})^n-n\int_{e^{-1}}^1(1+\ln x)^{n-1}\,dx\\ &=1-e^{-1}(1-1)^n-nI_{n-1}\\ &=1-nI_{n-1} \end{aligned}

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Calculus

Further IntegrationApply substitution, partial fractions, integration by parts and recurrence relationships to theoretical and practical problems.

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