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Q. 3

Expert-verifiedFound in: Page 417

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Explain why $\int \frac{2x}{{x}^{2}+1}dx$ and $\int \frac{1}{x\mathrm{ln}x}dx$ are essentially the same integral after a change of variables.

Both integrals turn into $\int \frac{1}{u}du$ after a change of variables; $u={x}^{2}+1$ in the first case, $u=\mathrm{ln}x$ in the second.

Explain why $\int \frac{2x}{{x}^{2}+1}dx$ and role="math" localid="1648740843374" $\int \frac{1}{x\mathrm{ln}x}dx$ are essentially the same integral after a change of variables.

Let ${x}^{2}+1=u$

$u={x}^{2}+1\phantom{\rule{0ex}{0ex}}\frac{du}{dx}=2x\phantom{\rule{0ex}{0ex}}du=2xdx$

This substitution changes the integral into

role="math" localid="1648740683170" $\int \frac{2x}{{x}^{2}+1}dx=\int \frac{1}{u}du$

Let $\mathrm{ln}x=u$

$u=\mathrm{ln}x\phantom{\rule{0ex}{0ex}}\frac{du}{dx}=\frac{1}{x}\phantom{\rule{0ex}{0ex}}du=\frac{1}{x}dx$

This substitution changes the integral into

localid="1648740902640" $\int \frac{1}{x\mathrm{ln}x}dx=\int \frac{1}{u}du$

Both integrals turn into $\int \frac{1}{u}du$ after a change of variables; $u={x}^{2}+1$ in the first case, $u=\mathrm{ln}x$ in the second.

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