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

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

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

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

ln x^{2} + 4 x + 8 + C

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= ln ∣ u ∣

Substitute back

u = x^{2} + 4 x + 8

= ln x^{2} + 4 x + 8

Add a constant to the solution

= ln x^{2} + 4 x + 8 + C

\int x + 2 d x

x d x + sin^{2} (x^{- 1} y) (y d x - x d y) = 0

- \frac{3}{2} x^{2} + 5 x + 5

tan (2 x - x^{3})

\frac{\partial}{\partial x} (x^{3} y + 2 x y^{3})

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