Get the answer to "Integration techniques. Use the methods introduced evaluate the..."

High School
Calculus and Analysis
Calculus 2
Applications of integrals

William Cleghorn

Answered question

2021-12-28

Integration techniques. Use the methods introduced evaluate the following integrals.

\int \frac{3 x}{\sqrt{x + 4}} d x

Answer & Explanation

vicki331g8

Beginner2021-12-29Added 37 answers

Step 1
Given integral is
$I = \int \frac{3 x}{\sqrt{x + 4}} d x$
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.
$\frac{1}{2 \sqrt{x + 4}} = \frac{d t}{d x}$
$\frac{d x}{\sqrt{x + 4}} = 2 d t$
Step 2
On substituting $x = t^{2} - 4$ and $\frac{d x}{\sqrt{x + 4}} = 2 d t$ in given integral, we get
$I = 3 \int (t^{2} - 4) 2 d t$
$= 6 \int (t^{2} - 4) d t + C$
$= 6 [\frac{t^{3}}{3} - 4 t] + C$
$= 2 t^{3} - 24 t + C$
Putting $t = \sqrt{x + 4}$ , we get
$I = 2 {(\sqrt{x + 4})}^{3} - 24 \sqrt{x + 4} + C$
$I = 2 {(\sqrt{x + 4})}^{\frac{3}{2}} - 24 \sqrt{x + 4} + C$
Step 3
Answer: The value of the given integral is $I = 2 {(\sqrt{x + 4})}^{\frac{3}{2}} - 24 \sqrt{x + 4} + C$ .

Thomas White

Beginner2021-12-30Added 40 answers

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

karton

Expert2022-01-04Added 613 answers

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

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Calculus 2

Didn't find what you were looking for?