$\text { Use integration tables to find } \int ( 2 x + 5 ) ^ { 2 } \sqrt { ( 2 x + 5 ) ^ { 2 } - 9 } d x \text {. }$
A) $\frac { 1 } { 32 } \left[ ( 2 x + 5 ) \left( 2 ( 2 x + 5 ) ^ { 2 } + 9 \right) \sqrt { ( 2 x + 5 ) ^ { 2 } - 9 } - 81 \ln \left| ( 2 x + 5 ) + \sqrt { ( 2 x + 5 ) ^ { 2 } + 9 } \right| \right] + C$
B) $\frac { 1 } { 16 } \left[ ( 2 x + 5 ) \left( 2 ( 2 x + 5 ) ^ { 2 } - 9 \right) \sqrt { ( 2 x + 5 ) ^ { 2 } - 9 } - 81 \ln \left| ( 2 x + 5 ) + \sqrt { ( 2 x + 5 ) ^ { 2 } - 9 } \right| \right] + C$
C) $\frac { 1 } { 16 } \left[ ( 2 x + 5 ) \left( 2 ( 2 x + 5 ) ^ { 2 } \pm 9 \right) \sqrt { ( 2 x + 5 ) ^ { 2 } \pm 9 } - 81 \ln \left| ( 2 x + 5 ) + \sqrt { ( 2 x + 5 ) ^ { 2 } \pm 9 } \right| \right] + C$
D)
$\frac { 1 } { 8 } \left[ ( 2 x + 5 ) \left( 2 ( 2 x + 5 ) ^ { 2 } - 9 \right) \sqrt { ( 2 x + 5 ) ^ { 2 } - 9 } - 81 \ln \left| ( 2 x + 5 ) + \sqrt { ( 2 x + 5 ) ^ { 2 } - 9 } \right| \right] + C$
E)
$\frac { 1 } { 16 } \left[ ( 2 x + 5 ) \left( 2 ( 2 x + 5 ) ^ { 2 } + 9 \right) \sqrt { ( 2 x + 5 ) ^ { 2 } + 9 } - 81 \ln \left| ( 2 x + 5 ) + \sqrt { ( 2 x + 5 ) ^ { 2 } + 9 } \right| \right] + C$

Question

$\text { Use integration tables to find } \int ( 2 x + 5 ) ^ { 2 } \sqrt { ( 2 x + 5 ) ^ { 2 } - 9 } d x \text {. }$ 
A) $\frac { 1 } { 32 } \left[ ( 2 x + 5 ) \left( 2 ( 2 x + 5 ) ^ { 2 } + 9 \right) \sqrt { ( 2 x + 5 ) ^ { 2 } - 9 } - 81 \ln \left| ( 2 x + 5 ) + \sqrt { ( 2 x + 5 ) ^ { 2 } + 9 } \right| \right] + C$
B) $\frac { 1 } { 16 } \left[ ( 2 x + 5 ) \left( 2 ( 2 x + 5 ) ^ { 2 } - 9 \right) \sqrt { ( 2 x + 5 ) ^ { 2 } - 9 } - 81 \ln \left| ( 2 x + 5 ) + \sqrt { ( 2 x + 5 ) ^ { 2 } - 9 } \right| \right] + C$
C) $\frac { 1 } { 16 } \left[ ( 2 x + 5 ) \left( 2 ( 2 x + 5 ) ^ { 2 } \pm 9 \right) \sqrt { ( 2 x + 5 ) ^ { 2 } \pm 9 } - 81 \ln \left| ( 2 x + 5 ) + \sqrt { ( 2 x + 5 ) ^ { 2 } \pm 9 } \right| \right] + C$
D)
$\frac { 1 } { 8 } \left[ ( 2 x + 5 ) \left( 2 ( 2 x + 5 ) ^ { 2 } - 9 \right) \sqrt { ( 2 x + 5 ) ^ { 2 } - 9 } - 81 \ln \left| ( 2 x + 5 ) + \sqrt { ( 2 x + 5 ) ^ { 2 } - 9 } \right| \right] + C$
E)
$\frac { 1 } { 16 } \left[ ( 2 x + 5 ) \left( 2 ( 2 x + 5 ) ^ { 2 } + 9 \right) \sqrt { ( 2 x + 5 ) ^ { 2 } + 9 } - 81 \ln \left| ( 2 x + 5 ) + \sqrt { ( 2 x + 5 ) ^ { 2 } + 9 } \right| \right] + C$

Quizplus · Accepted Answer

We can use the substitution $u = 2x + 5$, so $du = 2dx$. Then the integral becomes
$$\frac{1}{2}\int u^2 \sqrt{u^2-9}du.$$
We recognize this as a type of integral that can be solved using the substitution $u = 3\sec	heta$, so $du = 3\sec	heta	an	heta d	heta$. Making that substitution gives us
$$\frac{1}{2} \cdot 3^3\int	an^2	heta\cdot 3\sec	heta\cdot 3\sec	heta	an	heta d	heta = \frac{27}{2}\int	an^2	heta\sec	heta	an	heta d	heta.$$
We want to express this integral in terms of $\sec	heta$, so we can use the identity $	an^2	heta + 1 = \sec^2	heta$, or equivalently, $	an^2	heta = \sec^2	heta - 1$. Substituting and simplifying gives us
$$\frac{27}{2}\int(\sec^3	heta - \sec	heta)d	heta.$$
We can use integration by parts to evaluate this integral. Let $u = \sec	heta$ and $dv = \sec^2	heta d	heta$, so $du = \sec	heta	an	heta d	heta$ and $v = 	an	heta$. Then
$$\int\sec^3	heta d	heta = \sec	heta	an	heta - \int	an^2	heta\sec	heta d	heta = \sec	heta	an	heta - \int(\sec^2	heta-1)\sec	heta d	heta.$$
Simplifying and solving for the integral gives us
$$\int\sec^3	heta d	heta = \frac{1}{2}\left(\sec	heta	an	heta + \ln|\sec	heta+	an	heta|ight).$$
Using this result, we get
\begin{align*}
\int (2x + 5) ^ 2 \sqrt{(2x + 5) ^ 2 - 9 } dx &= \frac{1}{2}\int u^2\sqrt{u^2-9}du \
&= \frac{1}{2}\cdot\frac{27}{2}\int(\sec^3	heta - \sec	heta)d	heta \
&= \frac{27}{4}\left(\frac{1}{2}(2x+5)\sqrt{(2x+5)^2-9}+\frac{1}{2}\ln|(2x+5)+\sqrt{(2x+5)^2-9}|ight) + C \
&= \frac{1}{16}(2x+5)(2(2x+5)^2-9)\sqrt{(2x+5)^2-9}-\frac{81}{16}\ln|(2x+5)+\sqrt{(2x+5)^2-9}| + C.
\end{align*}
Therefore, the answer is B.

Use integration tables to find ∫(2x+5)2(2x+5)2−9dx. \text { Use integration tables to find } \int ( 2 x + 5 ) ^ { 2 } \sqrt { ( 2 x + 5 ) ^ { 2 } - 9 } d x \text {. } Use integration tables to find ∫(2x+5)2(2x+5)2−9​dx.

$\text { Use integration tables to find } \int ( 2 x + 5 ) ^ { 2 } \sqrt { ( 2 x + 5 ) ^ { 2 } - 9 } d x \text {. }$