What is the integral of (4x+8)/(sqrt(x^2+4x)) ?

The integral of (4x+8)/(sqrt(x^2+4x)) is 4sqrt(x^2+4x)+C

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integral of (4x+8)/(sqrt(x^2+4x))

\int \frac{4 x + 8}{x ^{2} + 4 x} d x

4 x^{2} + 4 x + C

Solution steps

\int \frac{4 x + 8}{x ^{2} + 4 x} d x

Expand

\frac{4 x + 8}{x ^{2} + 4 x} : \frac{4 x}{x ^{2} + 4 x} + \frac{8}{x ^{2} + 4 x}

= \int \frac{4 x}{x ^{2} + 4 x} + \frac{8}{x ^{2} + 4 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

= \int \frac{4 x}{x ^{2} + 4 x} d x + \int \frac{8}{x ^{2} + 4 x} d x

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

\int \frac{8}{x ^{2} + 4 x} d x = 8 ln (\frac{1}{2} x^{2} + 4 x + x + 2)

= 4 (x^{2} + 4 x - 2 ln (\frac{1}{2} x^{2} + 4 x + x + 2)) + 8 ln (\frac{1}{2} x^{2} + 4 x + x + 2)

Simplify

4 (x^{2} + 4 x - 2 ln (\frac{1}{2} x^{2} + 4 x + x + 2)) + 8 ln (\frac{1}{2} x^{2} + 4 x + x + 2) : 4 x^{2} + 4 x

= 4 x^{2} + 4 x

Add a constant to the solution

= 4 x^{2} + 4 x + C

y^{^{'}} = \frac{2 x ^{6} e ^{\frac{y}{x}} + x ^{4} y ^{2}}{x ^{5} y}

\frac{d}{d x} (2 x^{2} + 2 y - 8)

x \to + 1 + lim (\frac{2}{x} - \frac{1}{ln ( x )})

\frac{x ^{2}}{6 + x}

\frac{\partial}{\partial x} (x^{2} - y^{2} + z^{2} - 2 z - 4)

What is the integral of (4x+8)/(sqrt(x^2+4x)) ?
The integral of (4x+8)/(sqrt(x^2+4x)) is 4sqrt(x^2+4x)+C