LCC 206 {r3} integral rational expression

[karush]

$\tiny\text{LCC 206 {r3} integrarl rational expression}$
$$\displaystyle
I=\int \frac{3x}{\sqrt{x+4}}\,dx
=2\left(x-8\right)\sqrt{x+4}+C \\
\text{u substitution} \\
u=x+4 \ \ \ du=dx \ \ \ x=u-4 \\
\text{then} \\
I=3\int \frac{u-4}{\sqrt{u}} \ du
= 3\int {u}^{1/2}du -12\int {u}^{-1/2}\\
\text{integrate and back substitute } \\
I=\left[\frac{{6u}^{3/2}}{3}-\frac{{u}^{1/2}}{3}\right]
\implies 2\left(x+4\right)^{3/2}- 24\sqrt{x+4} \\
\text{simplify} \\
I= 2\left(x-8\right)\sqrt{x+4}+C \\
\text{not sure if trig substitution would have better?} $$
$\tiny\text
{from Surf the Nations math study group}$

 

[Prove It]

Re: LCC 206 {r3} integrarl rational expression

$\tiny\text{LCC 206 {r3} integrarl rational expression}$
$$\displaystyle
I=\int \frac{3x}{\sqrt{x+4}}\,dx
=2\left(x-8\right)\sqrt{x+4}+C \\
\text{u substitution} \\
u=x+4 \ \ \ du=dx \ \ \ x=u-4 \\
\text{then} \\
I=3\int \frac{u-4}{\sqrt{u}} \ du
= 3\int {u}^{1/2}du -12\int {u}^{-1/2}\\
\text{integrate and back substitute } \\
I=\left[\frac{{6u}^{3/2}}{3}-\frac{{u}^{1/2}}{3}\right]
\implies 2\left(x+4\right)^{3/2}- 24\sqrt{x+4} \\
\text{simplify} \\
I= 2\left(x-8\right)\sqrt{x+4}+C \\
\text{not sure if trig substitution would have better?} $$
$\tiny\text
{from Surf the Nations math study group}$

I think you have used the most concise method. Well done :)

 

[soroban]

$$\displaystyle
I=\int \frac{3x}{\sqrt{x+4}}\,dx
=2\left(x-8\right)\sqrt{x+4}+C \\
\text{u substitution} \\
u=x+4 \ \ \ du=dx \ \ \ x=u-4 \\
\text{then} \\
I=3\int \frac{u-4}{\sqrt{u}} \ du
= 3\int {u}^{1/2}du -12\int {u}^{-1/2}\\
\text{integrate and back substitute } \\
I=\left[\frac{{6u}^{3/2}}{3}-\frac{{u}^{1/2}}{3}\right]
\implies 2\left(x+4\right)^{3/2}- 24\sqrt{x+4} \\
\text{simplify} \\
I= 2\left(x-8\right)\sqrt{x+4}+C \\
\text{not sure if trig substitution would have better?} $$

Trig substitution is appropriate when we have functiions
with the forms: $$u^2 - a^2,\;u^2+a^2,\;a^2-u^2$$

If there is a linear function under the radical,
I often let $$u$$ equal the entire radical
.
$$\text{Let }u \;=\;\sqrt{x+4}\quad\Rightarrow\quad x \,=\,u^2-4 \quad\Rightarrow\quad du \,=\,2u\,du$$

$$\text{Substitute: }\;\int \frac{3(u^2-4)\cdot 2u\,du}{u} \;=\;6\int(u^2-4)\,du$$

. . . $$=\;\;6\left(\frac{u^3}{3} - 4u\right) + C \;\;=\;\;2u^3 - 24u + C \;\;=\;\;2u(u^2-12) + C$$

$$\text{Back-substitute: }\;2\sqrt{x+4}(x+4-12) + C \;\;=\;\; 2\sqrt{x+4}(x-8)+C$$

 

[Greg]

\(\displaystyle \int\dfrac{3x}{\sqrt{x+4}}\,dx\)

\(\displaystyle x=4\sinh^2(u),\quad u=\sinh^{-1}\left(\dfrac{\sqrt x}{2}\right)\)

\(\displaystyle dx=8\sinh(u)\cosh(u)\,du\)

\(\displaystyle \int\dfrac{3\cdot4\sinh^2(u)\cdot8\sinh(u)\cosh(u)\,du}{2\cosh(u)}\)

\(\displaystyle =48\int\sinh^3(u)\,du=48\int(\cosh^2(u)-1)\sinh(u)\,du\)

\(\displaystyle w=\cosh(u),\quad dw=\sinh(u)\,du\)

\(\displaystyle 48\int w^2-1\,dw=16w^3-48w+C\)

\(\displaystyle (*)\,\Leftrightarrow16w^3-48w\Rightarrow16\left(\dfrac{\sqrt{x+4}}{2}\right)^3-24\sqrt{x+4}+C\)

\(\displaystyle \int\dfrac{3x}{\sqrt{x+4}}\,dx=2(x-8)\sqrt{x+4}+C\)

$(*)$ Recall that $\cosh\left(\sinh^{-1}(z)\right)=\sqrt{z^2+1}$.