What is the integral from 0 to x of (t^2)/(1+t^3) ?

The integral from 0 to x of (t^2)/(1+t^3) is 1/3 ln|x^3+1|

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integral from 0 to x of (t^2)/(1+t^3)

\int_{0}^{x} \frac{t ^{2}}{1 + t ^{3}} d t

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

Solution steps

\int_{0}^{x} \frac{t ^{2}}{1 + t ^{3}} d t

Apply u-substitution

: \int_{1}^{1 + x^{3}} \frac{1}{3 u} d u

= \int_{1}^{1 + x^{3}} \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_{1}^{1 + x^{3}} \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3} [ln ∣ u ∣]_{1}^{1 + x^{3}}

Compute the boundaries:

ln x^{3} + 1

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

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What is the integral from 0 to x of (t^2)/(1+t^3) ?
The integral from 0 to x of (t^2)/(1+t^3) is 1/3 ln|x^3+1|