What is the integral of (t^2)/(at^3-b) ?

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

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integral of (t^2)/(at^3-b)

\int \frac{t ^{2}}{a t ^{3} - b} d t

\frac{1}{3 a} ln - b + a t^{3} + C

Solution steps

\int \frac{t ^{2}}{a t ^{3} - b} d t

Apply u-substitution

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

= \int \frac{1}{3 a u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3 a} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{3 a} ln ∣ u ∣

Substitute back

u = - b + a t^{3}

= \frac{1}{3 a} ln - b + a t^{3}

Add a constant to the solution

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

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