What is the integral of (2x)/(sqrt(x)+3) ?

The integral of (2x)/(sqrt(x)+3) is 4(1/3 x^{3/2}-(3x)/2+9sqrt(x)-27ln|sqrt(x)+3|)+C

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integral of (2x)/(sqrt(x)+3)

\int \frac{2 x}{x + 3} d x

4 (\frac{1}{3} x^{\frac{3}{2}} - \frac{3 x}{2} + 9 x - 27 ln x + 3) + C

Solution steps

\int \frac{2 x}{x + 3} d x

Take the constant out:

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

= 2 \cdot \int \frac{x}{x + 3} d x

Apply u-substitution

: \int \frac{2 u ^{3}}{u + 3} d u

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

Take the constant out:

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

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

Long division

\frac{u ^{3}}{u + 3} : u^{2} - 3 u + 9 - \frac{27}{u + 3}

= 2 \cdot 2 \cdot \int u^{2} - 3 u + 9 - \frac{27}{u + 3} 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

= 2 \cdot 2 (\int u^{2} d u - \int 3 u d u + \int 9 d u - \int \frac{27}{u + 3} d u)

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

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

\int 9 d u = 9 u

\int \frac{27}{u + 3} d u = 27 ln ∣ u + 3 ∣

= 2 \cdot 2 (\frac{u ^{3}}{3} - \frac{3 u ^{2}}{2} + 9 u - 27 ln ∣ u + 3 ∣)

Substitute back

u = x

= 2 \cdot 2 (\frac{( x ) ^{3}}{3} - \frac{3 ( x ) ^{2}}{2} + 9 x - 27 ln x + 3)

Simplify

2 \cdot 2 (\frac{( x ) ^{3}}{3} - \frac{3 ( x ) ^{2}}{2} + 9 x - 27 ln x + 3) : 4 (\frac{1}{3} x^{\frac{3}{2}} - \frac{3 x}{2} + 9 x - 27 ln x + 3)

= 4 (\frac{1}{3} x^{\frac{3}{2}} - \frac{3 x}{2} + 9 x - 27 ln x + 3)

Add a constant to the solution

= 4 (\frac{1}{3} x^{\frac{3}{2}} - \frac{3 x}{2} + 9 x - 27 ln x + 3) + C

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What is the integral of (2x)/(sqrt(x)+3) ?
The integral of (2x)/(sqrt(x)+3) is 4(1/3 x^{3/2}-(3x)/2+9sqrt(x)-27ln|sqrt(x)+3|)+C