William Cleghorn

2021-12-28

Integration techniques. Use the methods introduced evaluate the following integrals.
$\int \frac{3x}{\sqrt{x+4}}dx$

vicki331g8

Step 1
Given integral is
$I=\int \frac{3x}{\sqrt{x+4}}dx$
We will use the substitution method to solve the given integral.
Let $\sqrt{x+4}=t$,...(i)
Taking square on both sides.
$x+4={t}^{2}$
$x={t}^{2}-4$
Differentiate (i) with respect to x.
$\frac{1}{2\sqrt{x+4}}=\frac{dt}{dx}$
$\frac{dx}{\sqrt{x+4}}=2dt$
Step 2
On substituting $x={t}^{2}-4$ and $\frac{dx}{\sqrt{x+4}}=2dt$ in given integral, we get
$I=3\int \left({t}^{2}-4\right)2dt$
$=6\int \left({t}^{2}-4\right)dt+C$
$=6\left[\frac{{t}^{3}}{3}-4t\right]+C$
$=2{t}^{3}-24t+C$
Putting $t=\sqrt{x+4}$, we get
$I=2{\left(\sqrt{x+4}\right)}^{3}-24\sqrt{x+4}+C$
$I=2{\left(\sqrt{x+4}\right)}^{\frac{3}{2}}-24\sqrt{x+4}+C$
Step 3
Answer: The value of the given integral is $I=2{\left(\sqrt{x+4}\right)}^{\frac{3}{2}}-24\sqrt{x+4}+C$.

Thomas White

$\int \frac{3x}{\sqrt{x+4}}dx$
We make the change of variables:
$x+4={t}^{2}$
Therefore:
$x={t}^{2}-4$
dx=2tdt
$\int \frac{3\cdot {t}^{2}-12}{t}\cdot 2\cdot t\cdot dt$
Simplify the fractional expression:
$\int \left(6\cdot {t}^{2}-24\right)dt$
$\int \left(6\ast {t}^{2}-24\right)dt=2\ast {t}^{3}-24\ast t$
Substituting instead of $t=\sqrt{x+4}$, we get:
$I=2{\left(x+4\right)}^{\frac{3}{2}}-24\sqrt{x+4}+C$

karton

$\begin{array}{}\int \frac{3x}{\sqrt{x+4}}dx\\ \int \frac{3t-12}{\sqrt{t}}dt\\ \int \frac{3t-12}{{t}^{\frac{1}{2}}}dt\\ \int \frac{3t}{{t}^{\frac{1}{2}}}-\frac{12}{{t}^{\frac{1}{2}}}dt\\ \int 3{t}^{\frac{1}{2}}dt-\int \frac{12}{{t}^{\frac{1}{2}}}dt\\ 2t\sqrt{t}-24\sqrt{t}\\ 2\left(x+4\right)\sqrt{x+4}-24\sqrt{x+4}\\ 2\left(x+4\right)\sqrt{x+4}-24\sqrt{x+4}+C\end{array}$