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

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

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

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

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

Solution steps

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

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

Add a constant to the solution

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

\frac{d r}{d θ} = π sin (π θ), r (1) = 3

(- \frac{4}{3}) \cdot e^{- 2 t} + (- \frac{2}{3}) \cdot e^{t}

cos (t + 1)

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

\int \frac{1}{7 - 4 x} d x

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