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 \frac{3}{2 - x - x ^{2}} d x

\int_{0}^{\frac{π}{4}} cos (6 x) d x

y^{^{'}} (e^{x} - 1) ln (y) = y

- \frac{16}{( 8 x + 9 ) ^{3}}

\frac{d y}{d x} = \frac{y ^{2} + y}{x ^{2} + x}

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