What is the integral of (1x-2)/(sqrt(x)+4) ?

The integral of (1x-2)/(sqrt(x)+4) is -4x+2/3 x^{3/2}+28sqrt(x)-112ln|sqrt(x)+4|-16+C

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integral of (1x-2)/(sqrt(x)+4)

\int \frac{1 x - 2}{x + 4} d x

- 4 x + \frac{2}{3} x^{\frac{3}{2}} + 28 x - 112 ln x + 4 - 16 + C

Solution steps

\int \frac{1 \cdot x - 2}{x + 4} d x

Expand

\frac{1 x - 2}{x + 4} : \frac{1 \cdot x}{x + 4} - \frac{2}{x + 4}

Apply the Sum Rule:

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

= \int \frac{1 \cdot x}{x + 4} d x - \int \frac{2}{x + 4} d x

\int \frac{1 \cdot x}{x + 4} d x = 2 (\frac{1}{3} x^{\frac{3}{2}} - 2 x + 16 x - 64 ln x + 4)

\int \frac{2}{x + 4} d x = 4 (x + 4 - 4 ln x + 4)

= 2 (\frac{1}{3} x^{\frac{3}{2}} - 2 x + 16 x - 64 ln x + 4) - 4 (x + 4 - 4 ln x + 4)

Simplify

2 (\frac{1}{3} x^{\frac{3}{2}} - 2 x + 16 x - 64 ln x + 4) - 4 (x + 4 - 4 ln x + 4) : - 4 x + \frac{2}{3} x^{\frac{3}{2}} + 28 x - 112 ln x + 4 - 16

= - 4 x + \frac{2}{3} x^{\frac{3}{2}} + 28 x - 112 ln x + 4 - 16

Add a constant to the solution

= - 4 x + \frac{2}{3} x^{\frac{3}{2}} + 28 x - 112 ln x + 4 - 16 + C

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What is the integral of (1x-2)/(sqrt(x)+4) ?
The integral of (1x-2)/(sqrt(x)+4) is -4x+2/3 x^{3/2}+28sqrt(x)-112ln|sqrt(x)+4|-16+C