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

The integral of 1/(x(x^3+1)) is 1/3 (ln|x^3|-ln|1+x^3|)+C

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integral of 1/(x(x^3+1))

\int \frac{1}{x ( x ^{3} + 1 )} d x

\frac{1}{3} (ln x^{3} - ln 1 + x^{3}) + C

Solution steps

\int \frac{1}{x ( x ^{3} + 1 )} d x

Apply u-substitution

: \int \frac{1}{3 ( u - 1 ) u} d u

= \int \frac{1}{3 ( u - 1 ) u} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{( u - 1 ) u} d u

Take the partial fraction of

\frac{1}{( u - 1 ) u} : \frac{1}{u - 1} - \frac{1}{u}

= \frac{1}{3} \cdot \int \frac{1}{u - 1} - \frac{1}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (\int \frac{1}{u - 1} d u - \int \frac{1}{u} d u)

\int \frac{1}{u - 1} d u = ln ∣ u - 1 ∣

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{1}{3} (ln ∣ u - 1 ∣ - ln ∣ u ∣)

Substitute back

u = 1 + x^{3}

= \frac{1}{3} (ln 1 + x^{3} - 1 - ln 1 + x^{3})

Simplify

\frac{1}{3} (ln 1 + x^{3} - 1 - ln 1 + x^{3}) : \frac{1}{3} (ln x^{3} - ln 1 + x^{3})

= \frac{1}{3} (ln x^{3} - ln 1 + x^{3})

Add a constant to the solution

= \frac{1}{3} (ln x^{3} - ln 1 + x^{3}) + C

y = e^{3 x}, x = 0, x = 3

\frac{d}{d x} (e^{- 0.1 x})

f (x) = (x + 1)^{3} (x + 4) (x - 3)^{2}

x \to 3 + lim (x^{3} - c x)

f (x) = 4 x^{2} - 5 x, at x = - 1

What is the integral of 1/(x(x^3+1)) ?
The integral of 1/(x(x^3+1)) is 1/3 (ln|x^3|-ln|1+x^3|)+C