Integration techniques. Use the methods introduced evaluate the following integrals. ∫3xx+4dx

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

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

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