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

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

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

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

x^{2} + 4 x + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \frac{1}{2 u} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= \frac{1}{2} \cdot 2 u^{\frac{1}{2}}

Substitute back

u = x^{2} + 4 x

= \frac{1}{2} \cdot 2 (x^{2} + 4 x)^{\frac{1}{2}}

Simplify

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

= x^{2} + 4 x

Add a constant to the solution

= x^{2} + 4 x + C

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\int x^{2} e^{- 14 x} d x

\int_{5}^{8} 1

\frac{\partial}{\partial x} (cos^{7} (x^{6} y^{6}))

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