What is the integral of 1/(x^3+4x^2+8x) ?

The integral of 1/(x^3+4x^2+8x) is 1/8 ln|x|+1/16 (-ln| 1/4 x^2+x+2|-2arctan(1/2 x+1))+C

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integral of 1/(x^3+4x^2+8x)

\int \frac{1}{x ^{3} + 4 x ^{2} + 8 x} d x

\frac{1}{8} ln ∣ x ∣ + \frac{1}{16} (- ln \frac{1}{4} x^{2} + x + 2 - 2 arctan (\frac{1}{2} x + 1)) + C

Solution steps

\int \frac{1}{x ^{3} + 4 x ^{2} + 8 x} d x

Take the partial fraction of

\frac{1}{x ^{3} + 4 x ^{2} + 8 x} : \frac{1}{8 x} + \frac{- x - 4}{8 ( x ^{2} + 4 x + 8 )}

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

\int \frac{1}{8 x} d x = \frac{1}{8} ln ∣ x ∣

\int \frac{- x - 4}{8 ( x ^{2} + 4 x + 8 )} d x = \frac{1}{16} (- ln \frac{1}{4} x^{2} + x + 2 - 2 arctan (\frac{1}{2} x + 1))

= \frac{1}{8} ln ∣ x ∣ + \frac{1}{16} (- ln \frac{1}{4} x^{2} + x + 2 - 2 arctan (\frac{1}{2} x + 1))

Add a constant to the solution

= \frac{1}{8} ln ∣ x ∣ + \frac{1}{16} (- ln \frac{1}{4} x^{2} + x + 2 - 2 arctan (\frac{1}{2} x + 1)) + C

\int \frac{7 - 98 x}{4 - 49 x ^{2}} d x

e^{- 9 x}

x \to 1 + lim (\frac{7}{x - 1})

x ln (x) \frac{d y}{d x} + y = x e^{x}

\int ∣ x - 1 ∣ d x

What is the integral of 1/(x^3+4x^2+8x) ?
The integral of 1/(x^3+4x^2+8x) is 1/8 ln|x|+1/16 (-ln| 1/4 x^2+x+2|-2arctan(1/2 x+1))+C