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

The integral of (x^2ln^3(1+x^3))/(1+x^3) is 1/12 ln^4(1+x^3)+C

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integral of (x^2ln^3(1+x^3))/(1+x^3)

\int \frac{x ^{2} ln ^{3} ( 1 + x ^{3} )}{1 + x ^{3}} d x

\frac{1}{12} ln^{4} (1 + x^{3}) + C

Solution steps

\int \frac{x ^{2} ln ^{3} ( 1 + x ^{3} )}{1 + x ^{3}} d x

Apply u-substitution

: \int \frac{ln ^{3} ( u )}{3 u} d u

= \int \frac{ln ^{3} ( u )}{3 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{ln ^{3} ( u )}{u} d u

Apply u-substitution

: \int v^{3} d v

= \frac{1}{3} \cdot \int v^{3} d v

Apply the Power Rule

: \frac{v ^{4}}{4}

= \frac{1}{3} \cdot \frac{v ^{4}}{4}

Substitute back

= \frac{1}{3} \cdot \frac{ln ^{4} ( 1 + x ^{3} )}{4}

Simplify

\frac{1}{3} \cdot \frac{ln ^{4} ( 1 + x ^{3} )}{4} : \frac{1}{12} ln^{4} (1 + x^{3})

= \frac{1}{12} ln^{4} (1 + x^{3})

Add a constant to the solution

= \frac{1}{12} ln^{4} (1 + x^{3}) + C

x \to 0 lim (\frac{1}{2 ^{x}})

sin (tan (2 x))

\int 3 x (x^{2} - \frac{1}{3}) d x

f (x) = \frac{x}{x + 1}, at x = - 2

\frac{\partial}{\partial x} (- \frac{1}{2 x ^{2}})

What is the integral of (x^2ln^3(1+x^3))/(1+x^3) ?
The integral of (x^2ln^3(1+x^3))/(1+x^3) is 1/12 ln^4(1+x^3)+C