2012 H2 Mathematics Paper 1 Question 2

Integration Techniques

Answers

(i)

14ln⁡(1+x4)+C{\frac{1}{4} \ln \left( 1 + x^4 \right) + C}

(ii)

12tan⁡−1x2+C′{\frac{1}{2} \tan^{-1} x^2 + C'}

(iii)

0.186{0.186}

Full solutions

(i)

∫x31+x4  dx14∫4x31+x4  dx14ln⁡(1+x4)+C  ■\begin{align*} & \int \frac{x^3}{1 + x^4} \; \mathrm{d}x \\ & \frac{1}{4} \int \frac{4x^3}{1 + x^4} \; \mathrm{d}x \\ & \frac{1}{4} \ln \left( 1 + x^4 \right) + C \; \blacksquare\\ \end{align*}

(ii)

dudx=2x\frac{\mathrm{d}u}{\mathrm{d}x} = 2x
∫x1+x4  dx=∫x1+u212x  du=12∫11+u2  du=12tan⁡−1u+C′=12tan⁡−1x2+C′  ■\begin{align*} & \int \frac{x}{1 + x^4} \; \mathrm{d}x \\ & = \int \frac{x}{1+u^2} \frac{1}{2x} \; \mathrm{d}u \\ & = \frac{1}{2} \int \frac{1}{1+u^2} \; \mathrm{d}u \\ & = \frac{1}{2} \tan^{-1} u + C' \\ & = \frac{1}{2} \tan^{-1} x^2 + C' \; \blacksquare \end{align*}

(iii)

Using a GC,
∫01(x1+x4)2  dx=0.186 (3 dp)  ■\begin{align*} & \int_0^1 \left( \frac{x}{1+x^4} \right)^2 \; \mathrm{d}x \\ & = 0.186 \textrm{ (3 dp)} \; \blacksquare \\ \end{align*}