- #1
karush
Gold Member
MHB
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$\tiny\text{LCC 206 {r3} integrarl rational expression}$
$$\displaystyle
I=\int \frac{3x}{\sqrt{x+4}}\,dx
=2\left(x-8\right)\sqrt{x+4}+C \\
\text{u substitution} \\
u=x+4 \ \ \ du=dx \ \ \ x=u-4 \\
\text{then} \\
I=3\int \frac{u-4}{\sqrt{u}} \ du
= 3\int {u}^{1/2}du -12\int {u}^{-1/2}\\
\text{integrate and back substitute } \\
I=\left[\frac{{6u}^{3/2}}{3}-\frac{{u}^{1/2}}{3}\right]
\implies 2\left(x+4\right)^{3/2}- 24\sqrt{x+4} \\
\text{simplify} \\
I= 2\left(x-8\right)\sqrt{x+4}+C \\
\text{not sure if trig substitution would have better?} $$
$\tiny\text
{from Surf the Nations math study group}$
![Person surfing :person_surfing: 🏄](https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f3c4.png)
![Person surfing :person_surfing: 🏄](https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f3c4.png)
$$\displaystyle
I=\int \frac{3x}{\sqrt{x+4}}\,dx
=2\left(x-8\right)\sqrt{x+4}+C \\
\text{u substitution} \\
u=x+4 \ \ \ du=dx \ \ \ x=u-4 \\
\text{then} \\
I=3\int \frac{u-4}{\sqrt{u}} \ du
= 3\int {u}^{1/2}du -12\int {u}^{-1/2}\\
\text{integrate and back substitute } \\
I=\left[\frac{{6u}^{3/2}}{3}-\frac{{u}^{1/2}}{3}\right]
\implies 2\left(x+4\right)^{3/2}- 24\sqrt{x+4} \\
\text{simplify} \\
I= 2\left(x-8\right)\sqrt{x+4}+C \\
\text{not sure if trig substitution would have better?} $$
$\tiny\text
{from Surf the Nations math study group}$
![Person surfing :person_surfing: 🏄](https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f3c4.png)
![Person surfing :person_surfing: 🏄](https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f3c4.png)
![Person surfing :person_surfing: 🏄](https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f3c4.png)
![Person surfing :person_surfing: 🏄](https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f3c4.png)
![Person surfing :person_surfing: 🏄](https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f3c4.png)