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Q 41.

Expert-verifiedFound in: Page 429

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Solve the integral: $\int \frac{{x}^{2}+1}{{e}^{x}}dx$

The required answer is $-\frac{2x}{{e}^{x}}-\frac{{x}^{2}+1}{{e}^{x}}-2{e}^{-x}+C$.

We have given integral is $\int \frac{{x}^{2}+1}{{e}^{x}}dx$.

We have,

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

and

role="math" localid="1648720139610" $dv={e}^{-x}dx\phantom{\rule{0ex}{0ex}}v=\int {e}^{-x}dx\phantom{\rule{0ex}{0ex}}v=-\frac{1}{{e}^{x}}$

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

role="math" localid="1648720506794" $\int \frac{{x}^{2}+1}{{e}^{x}}dx\phantom{\rule{0ex}{0ex}}=\left[\left({x}^{2}+1\right)\left(-\frac{1}{{e}^{x}}\right)-\int \left(-\frac{1}{{e}^{x}}\right)dx\right]\phantom{\rule{0ex}{0ex}}=\left[-\frac{1}{{e}^{x}}\left({x}^{2}+1\right)-\int \left(-\frac{2x}{{e}^{x}}\right)dx\right]\phantom{\rule{0ex}{0ex}}=\left[-\frac{1}{{e}^{x}}\left({x}^{2}+1\right)+2\int \left(\frac{x}{{e}^{x}}\right)dx\right]\phantom{\rule{0ex}{0ex}}$

We have, $u=x\phantom{\rule{0ex}{0ex}}du=dx$

and $dv={e}^{-x}dx\phantom{\rule{0ex}{0ex}}v=\int {e}^{-x}dx\phantom{\rule{0ex}{0ex}}v=-\frac{1}{{e}^{x}}$

$==\left[-\frac{1}{{e}^{x}}\left({x}^{2}+1\right)+2\int \left({e}^{-x}\right)dx\right]\phantom{\rule{0ex}{0ex}}=\left[-\frac{1}{{e}^{x}}\left({x}^{2}+1\right)+2\left(x\left(-\frac{1}{{e}^{x}}\right)-\int \left(-\frac{1}{{e}^{x}}\right)dx\right)\right]\phantom{\rule{0ex}{0ex}}=\left[-\frac{1}{{e}^{x}}\left({x}^{2}+1\right)+2\left(-\frac{x}{{e}^{x}}-\int \left(-\frac{1}{{e}^{x}}\right)dx\right)\right]\phantom{\rule{0ex}{0ex}}=\left[-\frac{1}{{e}^{x}}\left({x}^{2}+1\right)+2\left(-\frac{x}{{e}^{x}}-\left(+{e}^{-x}\right)dx\right)\right]\phantom{\rule{0ex}{0ex}}=-\frac{2x}{{e}^{x}}-\frac{1}{{e}^{x}}\left({x}^{2}+1\right)-2{e}^{-x}+c$

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