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

Use the methods introduced evaluate the following integrals.
$$\displaystyle{\int_{{-{2}}}^{{{1}}}}{\frac{{{3}}}{{{x}^{{{2}}}+{4}{x}+{13}}}}{\left.{d}{x}\right.}$$

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Jordan Mitchell
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
$$\displaystyle{\int_{{-{2}}}^{{{1}}}}{\frac{{{3}}}{{{x}^{{{2}}}+{4}{x}+{13}}}}{\left.{d}{x}\right.}={3}{\int_{{{0}}}^{{{1}}}}{\frac{{{1}}}{{{3}{\left({v}^{{{2}}}+{1}\right)}}}}{\left.{d}{x}\right.}$$
$$\displaystyle{\int_{{{02}}}^{{{1}}}}{\frac{{{3}}}{{{x}^{{{2}}}+{4}{x}+{13}}}}{\left.{d}{x}\right.}={\int_{{{0}}}^{{{1}}}}{\frac{{{1}}}{{{\left({v}^{{{2}}}+{1}\right)}}}}{\left.{d}{x}\right.}$$
$$\displaystyle={{\left[{\arctan{{\left({v}\right)}}}\right]}_{{{0}}}^{{{1}}}}$$
$$\displaystyle={\frac{{\pi}}{{{4}}}}$$
###### Not exactly what youâ€™re looking for?
Terry Ray
$$\displaystyle{\int_{{-{2}}}^{{{1}}}}{\frac{{{3}}}{{{x}^{{{2}}}+{4}{x}+{13}}}}{\left.{d}{x}\right.}$$
We calculate the integral:
$$\displaystyle{3}\int{\frac{{{1}}}{{{\left({x}+{2}\right)}^{{{2}}}+{9}}}}{\left.{d}{x}\right.}={\arctan{{\left({\frac{{{1}}}{{{3}}}}\cdot{\left({x}+{2}\right)}\right)}}}$$
$$\displaystyle{\arctan{{\left({\frac{{{1}}}{{{3}}}}{\left({x}+{2}\right)}\right)}}}+{C}$$
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
$$\displaystyle{I}={\frac{{\pi}}{{{4}}}}-{\left({0}\right)}={\frac{{\pi}}{{{4}}}}$$
karton

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