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Integration techniques: JPJC 2024 Promo Q4

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Jurong Pioneer Junior College2024 PromoPromo Q4●○○ Foundational5 marks

What this question tests

Integration techniques.

Question

By writing $x=A(8x-8)+B$ , where $A$ and $B$ are constants, find $\displaystyle\int_0^1\frac{x}{4x^2-8x+5}\displaystyle\int x$ , giving your answer in the form $p\ln q+r\tan^{-1}s$ , where $p$ , $q$ , $r$ and $s$ are values to be determined. [5]

Show full worked solution▾

Consider $x=A(8x-8)+B$ .

By comparing coefficient of $x$ and constant term: $A=\dfrac18$ , $B=1$ .

$\begin{aligned} \int_0^1\frac{x}{4x^2-8x+5}\mathrm{d} x &=\frac18\int_0^1\frac{8x-8}{4x^2-8x+5}\mathrm{d} x +\int_0^1\frac1{4x^2-8x+5}\mathrm{d} x\\ &=\frac18\left[\ln(4x^2-8x+5)\right]_0^1 +\int_0^1\frac1{4[x^2-2x+(-1)^2-(-1)^2]+5}\mathrm{d} x\\ &=\frac18[\ln1-\ln5]+\int_0^1\frac1{4(x-1)^2+1}\mathrm{d} x\\ &=-\frac18\ln5+\int_0^1\frac1{4\left[(x-1)^2+\left(\frac12\right)^2\right]}\mathrm{d} x\\ &=-\frac18\ln5+\frac14\left[\frac1{\frac12}\tan^{-1}\left(\frac{x-1}{\frac12}\right)\right]_0^1\\ &=-\frac18\ln5+\frac12[\tan^{-1}(0)-\tan^{-1}(-2)]\\ &=-\frac18\ln5-\frac12\tan^{-1}(-2). \end{aligned}$

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