How do you use substitution to integrate $\frac{x}{\sqrt{x + 4}} d x$ $x/sqrt(x+4) dx$ ?

Question

How do you use substitution to integrate $\frac{x}{\sqrt{x + 4}} d x$ $x/sqrt(x+4) dx$ ?

2 Answers

Impact of this question

62788 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-29T17:29:38

The process is outlined below.

Explanation:

We are to evaluate, $\int \frac{x}{{(x + 4)}^{\frac{1}{2}}} \cdot d x$ $int x/(x + 4)^(1/2)*dx$ .

We shall accomplish this by substitution.

Let, $(x + 4) = t$ $(x + 4) = t$
$\Rightarrow d x = d t$ $implies dx = dt$ (By simple differentiation).

Thus, the integral becomes,

$\int \frac{x}{{(x + 4)}^{\frac{1}{2}}} \cdot d x = \int \frac{t - 4}{t^{\frac{1}{2}}} \cdot d t$ $int x/(x +4)^(1/2)*dx = int (t - 4)/t^(1/2)*dt$

$= \int t^{\frac{1}{2}} \cdot d t - \int 4 t^{- \frac{1}{2}} \cdot d t$ $= int t^(1/2)*dt - int 4t^(-1/2)*dt$

$= \frac{2}{3} t^{\frac{3}{2}} - 8 t^{\frac{1}{2}} + C$ $=2/3 t^(3/2) - 8t^(1/2) + C$ , where $C$ $C$ is the integration constant.

In terms of $x$ $x$ , the integral may be now written as,

$\int \frac{x}{{(x + 4)}^{\frac{1}{2}}} \cdot d x = \frac{2}{3} {(x + 4)}^{\frac{3}{2}} - 8 {(x + 4)}^{\frac{1}{2}} + C$ $int x/(x + 4)^(1/2)*dx =2/3 (x + 4)^(3/2) - 8(x + 4)^(1/2) + C$ .

Answer 2 · 2017-08-15T10:49:10

$\int \frac{x}{\sqrt{x + 4}} d x = \frac{2}{3} (x + 4) \sqrt{x + 4} - 8 \sqrt{x + 4} + C$ $int x/(sqrt(x+4)) "d"x = 2/3 (x+4)sqrt(x+4) - 8 sqrt(x+4) + C$

Explanation:

$\int \frac{x}{\sqrt{x + 4}} d x = \int \frac{x + 4 - 4}{\sqrt{x + 4}} d x$ $int x/(sqrt(x+4)) "d"x = int (x+4-4)/(sqrt(x+4)) "d"x$ ,
$\int \frac{x}{\sqrt{x + 4}} d x = \int \frac{x + 4}{{(x + 4)}^{\frac{1}{2}}} - \frac{4}{{(x + 4)}^{\frac{1}{2}}} d x$ $int x/(sqrt(x+4)) "d"x = int (x+4)/((x+4)^(1/2))-4/((x+4)^(1/2)) "d"x$ ,
$\int \frac{x}{\sqrt{x + 4}} d x = \int {(x + 4)}^{\frac{1}{2}} - 4 {(x + 4)}^{- \frac{1}{2}} d x$ $int x/(sqrt(x+4)) "d"x = int (x+4)^(1/2)-4(x+4)^(-1/2) "d"x$ ,
$\int \frac{x}{\sqrt{x + 4}} d x = \frac{2}{3} {(x + 4)}^{\frac{3}{2}} - 8 {(x + 4)}^{\frac{1}{2}} + C$ $int x/(sqrt(x+4)) "d"x = 2/3(x+4)^(3/2)-8(x+4)^(1/2) + C$

Science

Math

Humanities

... and beyond

How do you use substitution to integrate $\frac{x}{\sqrt{x + 4}} d x$ $x/sqrt(x+4) dx$ ?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you use substitution to integrate x√x+4dx x/sqrt(x+4) dx?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question

How do you use substitution to integrate $\frac{x}{\sqrt{x + 4}} d x$ $x/sqrt(x+4) dx$ ?