LCC 206 {review 7r5} Integral substitution}

[karush]

$\tiny\text{LCC 206 {review 7r5} Integral substitution}$
$$I=\int_{-2} ^{3} \frac{3}{{x}^{2}+4x+13}\,dx=
\arctan {\left(\frac{5}{3}\right)} $$
$\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} $
$\tiny\text{ Surf the Nations math study group}$

 

[MarkFL]

The formula is actually:

$$\int\frac{du}{u^2+a^2}=\frac{1}{a}\arctan\left(\frac{u}{a}\right)+C$$

And so you would have:

$$\left[\arctan\left(\frac{x+2}{3}\right)\right]_{-2}^3=\arctan {\left(\frac{5}{3}\right)}$$

 

[karush]

$\tiny\text{LCC 206 {review 7r5} Integral substitution}$

I just noticed the upper limit was $1$ so..

$$I=\int_{-2} ^{1} \frac{3}{{x}^{2}+4x+13}\,dx=
\frac{\pi}{4} $$
$\text{ complete the square } $
$$ \frac{3}{{x}^{2}+4x+13}
\implies\frac{3}{{\left(x+2 \right)}^{2}+{3}^{2}}$$
$\text{ then } $
$$\begin{align}\displaystyle
u& = x+2&
du& =dx&
a&= 3
\end{align}$$
$\text{ using the formula } $
$$\int\frac{du}{{u}^{2}+{a}^{2}}
=\arctan\left(\frac{u}{a}\right)+C$$
$\text{ results in } $
$$3\left[\arctan\left(\frac{x+2}{3}\right)\right]_{-2}^1
= \frac{\pi}{4}
$$

Where does 3 supposed to disappear?

$\tiny\text{ Surf the Nations math study group}$

 

[MarkFL]

...Where does 3 supposed to disappear?

The formula is actually:

$$\int\frac{du}{u^2+a^2}=\frac{1}{a}\arctan\left(\frac{u}{a}\right)+C$$