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

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

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

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

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

Solution steps

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

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 + t^{3}

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

Add a constant to the solution

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

\frac{d y}{d x}, x^{4} y - 8 x y + 3 x y^{2} = 9

x \to 0 lim (\frac{cos ( 3 x - \frac{π}{2} )}{x})

\frac{d}{d x} (ln (cot (5 x)))

f (x) = 4 x^{2} - 4 x + 1

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