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202104160738 Solution to 1970-AL-PMATH-II-8

(a) Find the values of the constants A, B so that

\displaystyle{\frac{-2x+4}{(x^2+1)(x-1)^2}=\frac{Ax+B}{x^2+1}-\frac{2}{x-1}+\frac{1}{(x-1)^2}} for all x.

Hence or otherwise, find the indefinite integral

\displaystyle{\int \frac{-2x+4}{(x^2+1)(x-1)^2}\,\mathrm{d}x}.

(b) Find the following indefinite integrals:

i. \displaystyle{\int x^5e^{x^3}}\,\mathrm{d}x;

ii. \displaystyle{\int \frac{\mathrm{d}x}{\sqrt{(x-a)(b-x)}}}.

Solution.

(a)

\begin{aligned} \textrm{RHS} & = \frac{(Ax+B)(x-1)^2-2(x^2+1)(x-1)+(x^2+1)}{(x^2+1)(x-1)^2} \\ & = \frac{[Ax^3+(B-2A)x^2+(A-2B)x+B]-[2(x^3-x^2+x-1)]+(x^2+1)}{(x^2+1)(x-1)^2} \\ & = \frac{(A-2)x^3+(B-2A+3)x^2+(A-2B-2)x+(B+3)}{(x^2+1)(x-1)^2} \\ & = \textrm{LHS} \end{aligned}

Comparing coefficients,

\begin{aligned} A-2 & = 0 \\ B-2A+3 & = 0 \\ A-2B-2 & = -2 \\ B+3 & = 4 \end{aligned}

one gets A=2 and B=1.

\begin{aligned} & \int \frac{-2x+4}{(x^2+1)(x-1)^2} \,\mathrm{d}x \\ = & \int \bigg( \frac{2x+1}{x^2+1} - \frac{2}{x-1} + \frac{1}{(x-1)^2} \bigg) \,\mathrm{d}x \\ = & \int \frac{2x+1}{x^2+1}\,\mathrm{d}x - \int \frac{2}{x-1}\,\mathrm{d}x + \int \frac{1}{(x-1)^2}\,\mathrm{d}x \end{aligned}

Evaluating term-by-term,  the first term

\begin{aligned} & \int \frac{2x+1}{x^2+1}\,\mathrm{d}x \\ \dots & \textrm{ Recall }1+\tan^2\theta = \sec^2\theta \enspace \dots \\ \dots & \textrm{ let }x=\tan\theta \enspace \dots \\ \dots & \textrm{ then }\mathrm{d}x = \sec^2\theta \,\mathrm{d}\theta \enspace \dots \\ = & \int \frac{2\tan\theta +1}{\tan^2\theta +1} \cdot \sec^2\theta \,\mathrm{d}\theta \\ = & \int \frac{2\tan\theta +1}{\sec^2\theta}\cdot \sec^2\theta\,\mathrm{d}\theta \\ = & \int (2\tan\theta +1)\,\mathrm{d}\theta \\ = & -2\ln |\cos\theta |+\theta +C_1 \\ = & 2\ln(\sqrt{x^2+1}) + \tan^{-1}x + C_1 \end{aligned}

the second term

\begin{aligned} & \int \frac{2}{x-1} \,\mathrm{d}x \\ & = \int \frac{2}{(x-1)}\,\mathrm{d}(x-1) \\ & = 2\ln (x-1) + C_2 \end{aligned}

and the third term

\begin{aligned} & \int\frac{1}{(x-1)^2}\,\mathrm{d}x \\ & = \int \frac{1}{(x-1)^2}\,\mathrm{d}(x-1) \\ & = -\frac{1}{(x-1)}+C_3 \end{aligned}

Therefore

\begin{aligned} & \displaystyle{\int \frac{-2x+4}{(x^2+1)(x-1)^2}\,\mathrm{d}x} \\ = & 2\ln \sqrt{x^2+1} + \tan^{-1}x - 2\ln (x-1) - \frac{1}{x-1} +C\\ = & \ln \Big( (x^2+1)/(x-1)^2 \Big) + \tan^{-1}x - \frac{1}{x-1} +C \end{aligned}

(b)

i.

\begin{aligned} & \int x^5e^{x^3}\,\mathrm{d}x \\ = & \int x^5e^{x^3}\cdot\frac{\mathrm{d}(x^3)}{3x^2} \\ = & \frac{1}{3}\int x^3e^{x^3}\,\mathrm{d}(x^3) \\ \dots\enspace & \textrm{let }u=x^3\enspace \dots \\ = & \frac{1}{3}\int ue^u\,\mathrm{d}u \end{aligned}

(to be continued)

Recall integration by parts from the product rule of differentiation

\begin{aligned} \frac{\mathrm{d}(uv)}{\mathrm{d}x} & = u\frac{\mathrm{d}v}{\mathrm{d}x} + v\frac{\mathrm{d}u}{\mathrm{d}x} \\ \int \frac{\mathrm{d}(uv)}{\mathrm{d}x}\,\mathrm{d}x & = \int u\bigg(\frac{\mathrm{d}v}{\mathrm{d}x}\bigg)\,\mathrm{d}x + \int v\bigg(\frac{\mathrm{d}u}{\mathrm{d}x}\bigg)\,\mathrm{d}x \\ \int \mathrm{d}(uv) & = \int u\,\mathrm{d}v + \int v\,\mathrm{d}u \\ \int u\,\mathrm{d}v & = uv - \int v\,\mathrm{d}u \end{aligned}

(continue)

\begin{aligned} = & \frac{1}{3} \int ue^u\,\mathrm{d}u \\ = & \frac{1}{3} \int u\,\mathrm{d}(e^u) \\ = & \frac{1}{3} \bigg( ue^u - \int e^u\,\mathrm{d}u \bigg) \\ = & \frac{1}{3}ue^u-\frac{1}{3}e^u + C \end{aligned}

ii. It is left the reader.

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