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Q 4 8.

Expert-verifiedFound in: Page 429

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Solve the integral: $\int x\mathrm{ln}\sqrt{x}dx$

The required answer is $\frac{{x}^{2}}{4}\mathrm{ln}\left(x\right)-\frac{{x}^{2}}{8}+c$.

We have given integral is $\int x\mathrm{ln}\sqrt{x}dx$.

We have,

$\int x\mathrm{ln}\sqrt{x}dx\phantom{\rule{0ex}{0ex}}=\int \left(\frac{x}{2}\mathrm{ln}x\right)dx\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\int x\mathrm{ln}xdx$

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

and

$dv=xdx\phantom{\rule{0ex}{0ex}}v=\int xdx\phantom{\rule{0ex}{0ex}}v=\frac{{x}^{2}}{2}$

The formula of integration by parts is $\int udv=uv-\int vdu$.

localid="1648797136253" $\frac{1}{2}\int x\mathrm{ln}xdx\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(\mathrm{ln}\left(x\right)\frac{{x}^{2}}{2}-\int \frac{{x}^{2}}{2}\frac{1}{x}dx\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(\frac{{x}^{2}}{2}\mathrm{ln}\left(x\right)-\frac{1}{2}\int xdx\right)\phantom{\rule{0ex}{0ex}}=\frac{{x}^{2}}{4}\mathrm{ln}\left(x\right)-\frac{{x}^{2}}{8}+c$

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