- #1
karush
Gold Member
MHB
- 3,269
- 5
$\tiny\text{LCC 206 {review 7r5} Integral substitution}$
$$I=\int_{-2} ^{3} \frac{3}{{x}^{2}+4x+13}\,dx=
\arctan {\left(\frac{5}{3}\right)} $$
$\text{ complete the square } $
$$ \frac{3}{{x}^{2}+4x+13}
\implies\frac{3}{{\left(x+2 \right)}^{2}+{3}^{2}}$$
$\text{ then } $
$$\begin{align}\displaystyle
u& = x+2 &
a&= 3 \\
\end{align}$$
$\text{ using the formula } $
$$\int\frac{dx}{{u}^{2}+{a}^{2}}
=\arctan\left(\frac{u}{a}\right)+C$$
$\text{ results in } $
$$3\left[\arctan\left(\frac{x+2}{3}\right)\right]_{-2}^3
= \arctan {\left(\frac{5}{3}\right)}
$$
$\text{saw the answer to my question in process
but thot would finish the post don't know another way to do it} $
$\tiny\text{ Surf the Nations math study group}$
$$I=\int_{-2} ^{3} \frac{3}{{x}^{2}+4x+13}\,dx=
\arctan {\left(\frac{5}{3}\right)} $$
$\text{ complete the square } $
$$ \frac{3}{{x}^{2}+4x+13}
\implies\frac{3}{{\left(x+2 \right)}^{2}+{3}^{2}}$$
$\text{ then } $
$$\begin{align}\displaystyle
u& = x+2 &
a&= 3 \\
\end{align}$$
$\text{ using the formula } $
$$\int\frac{dx}{{u}^{2}+{a}^{2}}
=\arctan\left(\frac{u}{a}\right)+C$$
$\text{ results in } $
$$3\left[\arctan\left(\frac{x+2}{3}\right)\right]_{-2}^3
= \arctan {\left(\frac{5}{3}\right)}
$$
$\text{saw the answer to my question in process
but thot would finish the post don't know another way to do it} $
$\tiny\text{ Surf the Nations math study group}$