Integration techniques: NJC 2024 Promo Q11

(Integration Techniques)

(a) (i) Let $$\frac{13x-2}{(3-x)(1+4x^2)}=\frac{A}{3-x}+\frac{Bx+C}{1+4x^2}.$$ Then $$13x-2=A(1+4x^2)+(Bx+C)(3-x).$$ Substitute $x=3$, $$\begin{aligned} 13(3)-2&=A[1+4(3)^2]\\ 37A&=37\\ A&=1. \end{aligned}$$ Substitute $x=0$ and $A=1$, $$\begin{aligned} -2&=1(1)+3C\\ 3C&=-3\\ C&=-1. \end{aligned}$$ Substitute $x=1$, $A=1$ and $C=-1$, $$\begin{aligned} 13(1)-2&=(1)[1+4(1)^2]+2(B-1)\\ 2(B-1)&=6\\ B&=4. \end{aligned}$$ Therefore, $$\frac{13x-2}{(3-x)(1+4x^2)}=\frac1{3-x}+\frac{4x-1}{1+4x^2}.$$ (a) (ii) $$\begin{aligned} \int_{-1/2}^{1/2}\frac{13x-2}{(3-x)(1+4x^2)}\,\mathrm{d} x &=\int_{-1/2}^{1/2}\left(\frac1{3-x}+\frac{4x-1}{1+4x^2}\right)\,\mathrm{d} x\\ &=\int_{-1/2}^{1/2}\frac1{3-x}\,\mathrm{d} x +\int_{-1/2}^{1/2}\frac{4x}{1+4x^2}\,\mathrm{d} x -\int_{-1/2}^{1/2}\frac1{1+4x^2}\,\mathrm{d} x\\ &=-\int_{-1/2}^{1/2}\frac1{3-x}\,\mathrm{d}(3-x) +\frac12\int_{-1/2}^{1/2}\frac{8x}{1+4x^2}\,\mathrm{d} x -\frac14\int_{-1/2}^{1/2}\frac1{(1/2)^2+x^2}\,\mathrm{d} x\\ &=\left[-\ln|3-x|+\frac12\ln(1+4x^2)-\frac12\tan^{-1}(2x)\right]_{-1/2}^{1/2}\\ &=-\ln\frac52+\ln\frac72+\frac12\ln2-\frac12\ln2 -\frac12[\tan^{-1}1-\tan^{-1}(-1)]. \end{aligned}$$

$$\begin{aligned} &=\ln\left(\frac{7/2}{5/2}\right)-\frac12\left(\frac\pi4+\frac\pi4\right)\\ &=\ln\frac75-\frac\pi4. \end{aligned}$$ (b) $x=a\tan t\Rightarrow\dfrac{\,\mathrm{d} x}{\,\mathrm{d} t}=a\sec^2t$.

$$\begin{aligned} \int\frac{a^2-x^2}{(a^2+x^2)^2}\,\mathrm{d} x &=\int\frac{a\sec^2t(a^2-a^2\tan^2t)}{(a^2+a^2\tan^2t)^2}\,\mathrm{d} t\\ &=\int\frac{a^3\sec^2t(1-\tan^2t)}{a^4(1+\tan^2t)^2}\,\mathrm{d} t\\ &=\int\frac{\sec^2t(1-\tan^2t)}{a\sec^4t}\,\mathrm{d} t\\ &=\frac1a\int(\cos^2t-\sin^2t)\,\mathrm{d} t\\ &=\frac1a\int\cos2t\,\mathrm{d} t\\ &=\frac1{2a}\sin2t+c\\ &=\frac1a\sin t\cos t+c\\ &=\frac1a\left(\frac{x}{\sqrt{a^2+x^2}}\right) \left(\frac{a}{\sqrt{a^2+x^2}}\right)+c\\ &=\frac{x}{a^2+x^2}+c, \end{aligned}$$ where $c$ is an arbitrary constant.