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

The integral of (x+2)/(x^2+4x) is 1/2 ln|x^2+4x|+C

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

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

\frac{1}{2} ln 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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = x^{2} + 4 x

= \frac{1}{2} ln x^{2} + 4 x

Add a constant to the solution

= \frac{1}{2} ln x^{2} + 4 x + C

y = 4^{- 5 x + 2}

\frac{d}{d x} (4 sin (x^{2}))

x y^{^{'}} - 2 y = - 2 x^{2} + x + 1

\frac{d y}{d t} + 8 t y = 5 t

\frac{d}{d x} (3 x + x^{2} - 3 x^{3} - x^{4})

What is the integral of (x+2)/(x^2+4x) ?
The integral of (x+2)/(x^2+4x) is 1/2 ln|x^2+4x|+C