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

The integral of (x^2ln(1+x^3))/(1+x^3) is 1/6 ln^2(1+x^3)+C

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

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

\frac{1}{6} ln^{2} (1 + x^{3}) + C

Solution steps

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

Apply u-substitution

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

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

\frac{ln ( 1 + u )}{1 + u} = ln (1 + u) \frac{1}{1 + u}

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

Apply u-substitution

: \int v d v

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

Apply the Power Rule

: \frac{v ^{2}}{2}

= \frac{1}{3} \cdot \frac{v ^{2}}{2}

Substitute back

= \frac{1}{3} \cdot \frac{ln ^{2} ( 1 + x ^{3} )}{2}

Simplify

\frac{1}{3} \cdot \frac{ln ^{2} ( 1 + x ^{3} )}{2} : \frac{1}{6} ln^{2} (1 + x^{3})

= \frac{1}{6} ln^{2} (1 + x^{3})

Add a constant to the solution

= \frac{1}{6} ln^{2} (1 + x^{3}) + C

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What is the integral of (x^2ln(1+x^3))/(1+x^3) ?
The integral of (x^2ln(1+x^3))/(1+x^3) is 1/6 ln^2(1+x^3)+C