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Original question & working

Reuse two basic integrals and a radical substitution · source reference

2025 TJC H2 Maths (9758) JC2 MSM · original exercise attribution retained on source pages · Question pp. 3 · Working pp. 13, 14

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Printed question

(a)(i) Find ∫2−x4x−x2−1dx\int\frac{2-x}{\sqrt{4x-x^2-1}}dx. [2]

(ii) Find ∫14x−x2−1dx\int\frac1{\sqrt{4x-x^2-1}}dx. [2]

(iii) Using (i) and (ii), find ∫x4x−x2−1dx\int\frac x{\sqrt{4x-x^2-1}}dx. [2]

(b)(i) By using u=xu=\sqrt x, find ∫x1+xdx\int\frac{\sqrt x}{1+\sqrt x}dx. [4]

(ii) Find the exact value of ∫09ln⁡(1+x)dx\int_0^9\ln(1+\sqrt x)dx, leaving the answer aln⁡b−c/ba\ln b-c/b, where a,b,ca,b,c are integers to be determined. [4]

[2016/VJC/Promo/11]

Printed working

(a)(i)

∫2−x4x−x2−1dx=12∫4−2x4x−x2−1dx=12(4x−x2−1)1/21/2+C=4x−x2−1+C.\int\frac{2-x}{\sqrt{4x-x^2-1}}dx=\frac12\int\frac{4-2x}{\sqrt{4x-x^2-1}}dx=\frac12\frac{(4x-x^2-1)^{1/2}}{1/2}+C=\sqrt{4x-x^2-1}+C.

(a)(ii)

∫dx4x−x2−1=∫dx3−(x−2)2=∫1 dx(3)2−(x−2)2=sin⁡−1x−23+C.\int\frac{dx}{\sqrt{4x-x^2-1}}=\int\frac{dx}{\sqrt{3-(x-2)^2}}=\int\frac{1\,dx}{\sqrt{(\sqrt3)^2-(x-2)^2}}=\sin^{-1}\frac{x-2}{\sqrt3}+C.

(a)(iii)

∫x4x−x2−1dx=−∫−x+2−24x−x2−1dx=−∫2−x4x−x2−1dx+∫24x−x2−1dx=−4x−x2−1+2sin⁡−1x−23+C.\int\frac x{\sqrt{4x-x^2-1}}dx=-\int\frac{-x+2-2}{\sqrt{4x-x^2-1}}dx=-\int\frac{2-x}{\sqrt{4x-x^2-1}}dx+\int\frac2{\sqrt{4x-x^2-1}}dx=-\sqrt{4x-x^2-1}+2\sin^{-1}\frac{x-2}{\sqrt3}+C.

(b)(i)

u=xu=\sqrt x, du/dx=1/(2x)du/dx=1/(2\sqrt x), dx/du=2x=2udx/du=2\sqrt x=2u. Alternatively u2=xu^2=x, 2u du/dx=12u\,du/dx=1, dx/du=2udx/du=2u.

∫x1+xdx=∫u1+u(2u du)=∫2u21+udu=∫(2u−2+21+u)du=u2−2u+2ln⁡∣1+u∣+C=x−2x+2ln⁡(1+x)+C,\int\frac{\sqrt x}{1+\sqrt x}dx=\int\frac u{1+u}(2u\,du)=\int\frac{2u^2}{1+u}du=\int\left(2u-2+\frac2{1+u}\right)du=u^2-2u+2\ln|1+u|+C=x-2\sqrt x+2\ln(1+\sqrt x)+C,

since 1+x>01+\sqrt x>0.

(b)(ii)

∫09ln⁡(1+x)dx=[xln⁡(1+x)]09−∫09x11+x12xdx=9ln⁡4−12∫09x1+xdx=9ln⁡4−12[x−2x+2ln⁡(1+x)]09=9ln⁡4−12(9−29+2ln⁡4)=16ln⁡2−32.\begin{aligned}\int_0^9\ln(1+\sqrt x)dx&=[x\ln(1+\sqrt x)]_0^9-\int_0^9x\frac1{1+\sqrt x}\frac1{2\sqrt x}dx\\&=9\ln4-\frac12\int_0^9\frac{\sqrt x}{1+\sqrt x}dx\\&=9\ln4-\frac12[x-2\sqrt x+2\ln(1+\sqrt x)]_0^9\\&=9\ln4-\frac12(9-2\sqrt9+2\ln4)=16\ln2-\frac32.\end{aligned}

Teacher notes

(a)(i) power formula; (a)(ii) MF27 ∫dx/a2−x2=sin⁡−1(x/a)+c\int dx/\sqrt{a^2-x^2}=\sin^{-1}(x/a)+c, and the not-in-MF27 composite formula ∫f′(x)/a2−(f(x))2dx=sin⁡−1(f(x)/a)+c\int f'(x)/\sqrt{a^2-(f(x))^2}dx=\sin^{-1}(f(x)/a)+c, where f=x−2,a=3f=x-2,a=\sqrt3.

(b)(i) substitution ∫f(x)dx=∫g(u)(dx/du)du\int f(x)dx=\int g(u)(dx/du)du; final answer of an indefinite integral in xx must be in xx. (b)(ii) parts with u=ln⁡(1+x)u=\ln(1+\sqrt x), dv/dx=1dv/dx=1.

Printed answer key

(a)(i) 4x−x2−1+C\sqrt{4x-x^2-1}+C; (ii) sin⁡−1((x−2)/3)+C\sin^{-1}((x-2)/\sqrt3)+C; (iii) −[4x−x2−1]+2sin⁡−1((x−2)/3)+C-[\sqrt{4x-x^2-1}]+2\sin^{-1}((x-2)/\sqrt3)+C. (b)(i) x−2x+2ln⁡∣1+x∣+Cx-2\sqrt x+2\ln|1+\sqrt x|+C; (ii) 16ln⁡2−3/216\ln2-3/2.