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

hadejada7x 2022-01-04 Answered
Use the methods introduced evaluate the following integrals.
\(\displaystyle{\int_{{-{2}}}^{{{1}}}}{\frac{{{3}}}{{{x}^{{{2}}}+{4}{x}+{13}}}}{\left.{d}{x}\right.}\)

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Expert Answer

Jordan Mitchell
Answered 2022-01-05 Author has 2657 answers
Step 1
The given integral is \(\displaystyle{\int_{{-{2}}}^{{{1}}}}{\frac{{{3}}}{{{x}^{{{2}}}+{4}{x}+{13}}}}{\left.{d}{x}\right.}\).
Rewrite the given integral as follows:
\(\displaystyle{\int_{{-{2}}}^{{{1}}}}{\frac{{{3}}}{{{x}^{{{2}}}+{4}{x}+{13}}}}{\left.{d}{x}\right.}={3}{\int_{{-{2}}}^{{{1}}}}{\frac{{{1}}}{{{x}^{{{2}}}+{4}{x}+{13}}}}{\left.{d}{x}\right.}\)
\(\displaystyle={3}{\int_{{-{2}}}^{{{1}}}}{\frac{{{1}}}{{{\left({x}+{2}\right)}^{{{2}}}+{9}}}}\)
Step 2
Apply u-substitution:
\(\displaystyle{u}={x}+{2}\Rightarrow{d}{u}={\left.{d}{x}\right.}\)
when \(\displaystyle{x}={1}\Rightarrow{u}={3}\)
when \(\displaystyle{x}=-{2}\Rightarrow{u}={0}\)
Now,
\(\displaystyle{\int_{{-{2}}}^{{{1}}}}{\frac{{{3}}}{{{x}^{{{2}}}+{4}{x}+{13}}}}{\left.{d}{x}\right.}={3}{\int_{{{0}}}^{{{3}}}}{\frac{{{1}}}{{{u}^{{{2}}}+{9}}}}{\left.{d}{x}\right.}\)
Apply integral substitution: u=3v
\(\displaystyle{\int_{{-{2}}}^{{{1}}}}{\frac{{{3}}}{{{x}^{{{2}}}+{4}{x}+{13}}}}{\left.{d}{x}\right.}={3}{\int_{{{0}}}^{{{1}}}}{\frac{{{1}}}{{{3}{\left({v}^{{{2}}}+{1}\right)}}}}{\left.{d}{x}\right.}\)
\(\displaystyle{\int_{{{02}}}^{{{1}}}}{\frac{{{3}}}{{{x}^{{{2}}}+{4}{x}+{13}}}}{\left.{d}{x}\right.}={\int_{{{0}}}^{{{1}}}}{\frac{{{1}}}{{{\left({v}^{{{2}}}+{1}\right)}}}}{\left.{d}{x}\right.}\)
\(\displaystyle={{\left[{\arctan{{\left({v}\right)}}}\right]}_{{{0}}}^{{{1}}}}\)
\(\displaystyle={\frac{{\pi}}{{{4}}}}\)
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Terry Ray
Answered 2022-01-06 Author has 5276 answers
\(\displaystyle{\int_{{-{2}}}^{{{1}}}}{\frac{{{3}}}{{{x}^{{{2}}}+{4}{x}+{13}}}}{\left.{d}{x}\right.}\)
We calculate the integral:
\(\displaystyle{3}\int{\frac{{{1}}}{{{\left({x}+{2}\right)}^{{{2}}}+{9}}}}{\left.{d}{x}\right.}={\arctan{{\left({\frac{{{1}}}{{{3}}}}\cdot{\left({x}+{2}\right)}\right)}}}\)
Answer:
\(\displaystyle{\arctan{{\left({\frac{{{1}}}{{{3}}}}{\left({x}+{2}\right)}\right)}}}+{C}\)
We calculate the definite integral:
\(\displaystyle{\int_{{-{2}}}^{{{1}}}}{\frac{{{3}}}{{{x}^{{{2}}}+{4}{x}+{13}}}}{\left.{d}{x}\right.}={\left({\arctan{{\left({\frac{{{1}}}{{{3}}}}\cdot{\left({x}+{2}\right)}\right)}}}\right)}{{\mid}_{{-{2}}}^{{{1}}}}\)
\(\displaystyle{F}{\left({1}\right)}={\frac{{\pi}}{{{4}}}}\)
F(-2)=0
\(\displaystyle{I}={\frac{{\pi}}{{{4}}}}-{\left({0}\right)}={\frac{{\pi}}{{{4}}}}\)
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karton
Answered 2022-01-11 Author has 8659 answers

\(\int_{-2}^{1}\frac{3}{x^{2}+4x+13}dx \\\int \frac{3}{x^{2}+4x+13}dx \\\int \frac{3}{x^{2}+4x+4+9}dx \\\int \frac{3}{(x+2)^{2}+9}dx \\3\times \int \frac{1}{(x+2)^{2}+9}dx \\3\times \int \frac{1}{t^{2}+9}dt \\3\times \frac{1}{3}\times \arctan(\frac{t}{3}) \\3\times \frac{1}{3}\times \arctan (\frac{x+2}{3}) \\\arctan(\frac{x+2}{3}) \\\arctan(\frac{x+2}{3})|_{-2}^{1} \\\arctan(\frac{1+2}{3})-\arctan(\frac{-2+2}{3}) \\Answer: \\\frac{\pi}{4}\)

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