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

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

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

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

\frac{1}{3} ln 1 + x^{3} + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{3 u} d u

= \int \frac{1}{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{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} ln ∣ u ∣

Substitute back

u = 1 + x^{3}

= \frac{1}{3} ln 1 + x^{3}

Add a constant to the solution

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

x \to - 1 lim (3 x^{2} - 2 x - 1)

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

f (x) = 8 x^{2} + 4 x - 6, at x = - 3

\int \frac{u - 2 u ^{2}}{u} d u

\int a e^{x} d x

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