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

The integral of t(2-3t^2)^2 is -1/18 (2-3t^2)^3+C

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integral of t(2-3t^2)^2

\int t (2 - 3 t^{2})^{2} d t

- \frac{1}{18} (2 - 3 t^{2})^{3} + C

Solution steps

\int t (2 - 3 t^{2})^{2} d t

Apply u-substitution

: \int - \frac{u ^{2}}{6} d u

= \int - \frac{u ^{2}}{6} d u

Take the constant out:

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

= - \frac{1}{6} \cdot \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= - \frac{1}{6} \cdot \frac{u ^{3}}{3}

Substitute back

u = 2 - 3 t^{2}

= - \frac{1}{6} \cdot \frac{( 2 - 3 t ^{2} ) ^{3}}{3}

Simplify

- \frac{1}{6} \cdot \frac{( 2 - 3 t ^{2} ) ^{3}}{3} : - \frac{1}{18} (2 - 3 t^{2})^{3}

= - \frac{1}{18} (2 - 3 t^{2})^{3}

Add a constant to the solution

= - \frac{1}{18} (2 - 3 t^{2})^{3} + C

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What is the integral of t(2-3t^2)^2 ?
The integral of t(2-3t^2)^2 is -1/18 (2-3t^2)^3+C