Solutions - 0060-09

(a) (1) $u$-substitution with $u=e^x$ and $du=e^{x}dx$:

Notice that $dx=u^{-1}du$. Then:

$$\begin{array}{c}\int \frac{1}{e^x-1}dx\gg \gg \int \frac{1}{u-1}{u}^{-1}du\\ \\ \gg \gg \int \frac{1}{u(u-1)}du\end{array}$$

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(2) Partial fractions:

$$\begin{array}{c}\frac{1}{u(u-1)}=\frac{A}{u}+\frac{B}{u-1}\\ \\ \gg \gg 1=A(u-1)+B(u)\\ \\ \text{set }u=1:\gg \gg 1=B\\ \\ \text{set }u=0:\gg \gg 1=A(-1)\gg \gg A=-1\end{array}$$

Therefore:

$$\begin{array}{c}\int \frac{1}{u(u-1)}du\gg \gg -\int \frac{1}{u}du+\int \frac{1}{u-1}du\\ \\ \gg \gg -\ln |u|+\ln |u-1|+C\\ \\ \gg \gg -x+\ln |e^x-1|+C\end{array}$$

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(b) (1) Set $u=\sqrt{x}$, so $x=u^2$ and $dx=2udu$:

$$\int \frac{\sqrt{x}}{x-1}dx\gg \gg \int \frac{2u^2}{u^2-1}du$$

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(2) Partial fractions:

$$\begin{array}{c}\frac{2u^2}{u^2-1}\gg \gg \frac{2u^2}{(u+1)(u-1)}=\frac{A}{u+1}+\frac{B}{u-1}\\ \\ \gg \gg 2u^2=A(u-1)+B(u+1)\\ \\ \text{set }u=1:\gg \gg 2=B(2)\gg \gg B=1\\ \\ \text{set }u=-1:\gg \gg 2=A(-2)\gg \gg A=-1\end{array}$$

Therefore:

$$\begin{array}{c}\int \frac{2u^2}{u^2-1}du\gg \gg \int \frac{1}{u+1}du+\int \frac{-1}{u-1}du\\ \\ \gg \gg \ln |u+1|-\ln |u-1|+C\\ \\ \gg \gg \ln \left|\frac{u+1}{u-1}\right|+C\\ \\ \gg \gg \ln \left|\frac{\sqrt{x}+1}{\sqrt{x}-1}\right|+C\end{array}$$