What is the integral of 3/(sqrt(x+3)+\sqrt{x)} ?

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

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integral of 3/(sqrt(x+3)+\sqrt{x)}

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

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

Solution steps

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

Take the constant out:

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

= 3 \cdot \int \frac{1}{x + 3 + x} d x

Multiply by the conjugate of

x + 3 + x : \frac{3}{x + 3 - x}

= 3 \cdot \int \frac{1}{\frac{3}{x + 3 - x}} d x

Take the constant out:

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

= 3 \cdot \frac{1}{3} \cdot \int \frac{1}{\frac{1}{x + 3 - x}} d x

Simplify

= 3 \cdot \frac{1}{3} \cdot \int x + 3 - x d x

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 \frac{1}{3} (\int x + 3 d x - \int x d x)

\int x + 3 d x = \frac{2}{3} (x + 3)^{\frac{3}{2}}

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d t} + y = 7 t

(7, - 2), (3, 5)

x \to 0 lim (\frac{7 x ^{2} - 6 x}{x})

\frac{d}{d x} (8 x^{3} - cos (4 x))

\int x + \frac{1}{x} - 7 d x

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