What is the integral of (3x)/(sqrt(x+4)) ?

The integral of (3x)/(sqrt(x+4)) is 6(1/3 (x+4)^{3/2}-4sqrt(x+4))+C

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integral of (3x)/(sqrt(x+4))

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

6 (\frac{1}{3} (x + 4)^{\frac{3}{2}} - 4 x + 4) + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int \frac{x}{x + 4} d x

Apply u-substitution

: \int 2 (u^{2} - 4) d u

= 3 \cdot \int 2 (u^{2} - 4) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot 2 \cdot \int u^{2} - 4 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 \cdot 2 (\int u^{2} d u - \int 4 d u)

\int u^{2} d u = \frac{u ^{3}}{3}

\int 4 d u = 4 u

= 3 \cdot 2 (\frac{u ^{3}}{3} - 4 u)

Substitute back

u = x + 4

= 3 \cdot 2 (\frac{( x + 4 ) ^{3}}{3} - 4 x + 4)

Simplify

3 \cdot 2 (\frac{( x + 4 ) ^{3}}{3} - 4 x + 4) : 6 (\frac{1}{3} (x + 4)^{\frac{3}{2}} - 4 x + 4)

= 6 (\frac{1}{3} (x + 4)^{\frac{3}{2}} - 4 x + 4)

Add a constant to the solution

= 6 (\frac{1}{3} (x + 4)^{\frac{3}{2}} - 4 x + 4) + C

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What is the integral of (3x)/(sqrt(x+4)) ?
The integral of (3x)/(sqrt(x+4)) is 6(1/3 (x+4)^{3/2}-4sqrt(x+4))+C