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

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

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

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

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

Solution steps

\int \frac{1}{x ( 1 + x ^{3} )} 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 = 5 x, y = 4, 2 y + x = 6

\int_{0}^{1} x 1 + x^{2} d x

x^{6} (1 - \frac{1}{x + 5})

\int_{1}^{8} \frac{7 ln ( x )}{x ^{2}} d x

x \to 2 lim (\frac{1}{x} - \frac{1}{x ^{2}})

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