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

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36
Jul 29, 2015

The process is outlined below.

#### Explanation:

We are to evaluate, $\int \frac{x}{x + 4} ^ \left(\frac{1}{2}\right) \cdot \mathrm{dx}$.

We shall accomplish this by substitution.

Let, $\left(x + 4\right) = t$
$\implies \mathrm{dx} = \mathrm{dt}$ (By simple differentiation).

Thus, the integral becomes,

$\int \frac{x}{x + 4} ^ \left(\frac{1}{2}\right) \cdot \mathrm{dx} = \int \frac{t - 4}{t} ^ \left(\frac{1}{2}\right) \cdot \mathrm{dt}$

$= \int {t}^{\frac{1}{2}} \cdot \mathrm{dt} - \int 4 {t}^{- \frac{1}{2}} \cdot \mathrm{dt}$

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

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

$\int \frac{x}{x + 4} ^ \left(\frac{1}{2}\right) \cdot \mathrm{dx} = \frac{2}{3} {\left(x + 4\right)}^{\frac{3}{2}} - 8 {\left(x + 4\right)}^{\frac{1}{2}} + C$.

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10
Aug 15, 2017

$\int \frac{x}{\sqrt{x + 4}} \text{d} x = \frac{2}{3} \left(x + 4\right) \sqrt{x + 4} - 8 \sqrt{x + 4} + C$

#### Explanation:

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

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#### Explanation

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#### Explanation:

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5
Aug 11, 2017

Let ${u}^{2} = x + 4$
Then $2 u \cdot \mathrm{du} = \mathrm{dx}$
$x = {u}^{2} - 4$
Now the integral becomes
$\int \frac{{u}^{2} - 4}{u} \cdot 2 u \cdot \mathrm{du}$
= $2 \int \left({u}^{2} - 4\right) \mathrm{du}$
= $2 \left({u}^{3} / 3 - 4 u\right)$
= $\frac{2}{3} {\left(x + 4\right)}^{\frac{3}{2}} - 8 \sqrt{x + 4}$

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