2022-01-04

Use the methods introduced evaluate the following integrals.
${\int }_{-2}^{1}\frac{3}{{x}^{2}+4x+13}dx$

Jordan Mitchell

Step 1
The given integral is ${\int }_{-2}^{1}\frac{3}{{x}^{2}+4x+13}dx$.
Rewrite the given integral as follows:
${\int }_{-2}^{1}\frac{3}{{x}^{2}+4x+13}dx=3{\int }_{-2}^{1}\frac{1}{{x}^{2}+4x+13}dx$
$=3{\int }_{-2}^{1}\frac{1}{{\left(x+2\right)}^{2}+9}$
Step 2
Apply u-substitution:
$u=x+2⇒du=dx$
when $x=1⇒u=3$
when $x=-2⇒u=0$
Now,
${\int }_{-2}^{1}\frac{3}{{x}^{2}+4x+13}dx=3{\int }_{0}^{3}\frac{1}{{u}^{2}+9}dx$
Apply integral substitution: u=3v
${\int }_{-2}^{1}\frac{3}{{x}^{2}+4x+13}dx=3{\int }_{0}^{1}\frac{1}{3\left({v}^{2}+1\right)}dx$
${\int }_{02}^{1}\frac{3}{{x}^{2}+4x+13}dx={\int }_{0}^{1}\frac{1}{\left({v}^{2}+1\right)}dx$
$={\left[\mathrm{arctan}\left(v\right)\right]}_{0}^{1}$
$=\frac{\pi }{4}$

Terry Ray

${\int }_{-2}^{1}\frac{3}{{x}^{2}+4x+13}dx$
We calculate the integral:
$3\int \frac{1}{{\left(x+2\right)}^{2}+9}dx=\mathrm{arctan}\left(\frac{1}{3}\cdot \left(x+2\right)\right)$
$\mathrm{arctan}\left(\frac{1}{3}\left(x+2\right)\right)+C$
We calculate the definite integral:
${\int }_{-2}^{1}\frac{3}{{x}^{2}+4x+13}dx=\left(\mathrm{arctan}\left(\frac{1}{3}\cdot \left(x+2\right)\right)\right){\mid }_{-2}^{1}$
$F\left(1\right)=\frac{\pi }{4}$
F(-2)=0
$I=\frac{\pi }{4}-\left(0\right)=\frac{\pi }{4}$

karton

${\int }_{-2}^{1}\frac{3}{{x}^{2}+4x+13}dx\phantom{\rule{0ex}{0ex}}\int \frac{3}{{x}^{2}+4x+13}dx\phantom{\rule{0ex}{0ex}}\int \frac{3}{{x}^{2}+4x+4+9}dx\phantom{\rule{0ex}{0ex}}\int \frac{3}{\left(x+2{\right)}^{2}+9}dx\phantom{\rule{0ex}{0ex}}3×\int \frac{1}{\left(x+2{\right)}^{2}+9}dx\phantom{\rule{0ex}{0ex}}3×\int \frac{1}{{t}^{2}+9}dt\phantom{\rule{0ex}{0ex}}3×\frac{1}{3}×\mathrm{arctan}\left(\frac{t}{3}\right)\phantom{\rule{0ex}{0ex}}3×\frac{1}{3}×\mathrm{arctan}\left(\frac{x+2}{3}\right)\phantom{\rule{0ex}{0ex}}\mathrm{arctan}\left(\frac{x+2}{3}\right)\phantom{\rule{0ex}{0ex}}\mathrm{arctan}\left(\frac{x+2}{3}\right){|}_{-2}^{1}\phantom{\rule{0ex}{0ex}}\mathrm{arctan}\left(\frac{1+2}{3}\right)-\mathrm{arctan}\left(\frac{-2+2}{3}\right)\phantom{\rule{0ex}{0ex}}Answer:\phantom{\rule{0ex}{0ex}}\frac{\pi }{4}$

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