What is the integral of 12t(3t^2-1)^3 ?

The integral of 12t(3t^2-1)^3 is 1/2 (3t^2-1)^4+C

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integral of 12t(3t^2-1)^3

\int 12 t (3 t^{2} - 1)^{3} d t

\frac{1}{2} (3 t^{2} - 1)^{4} + C

Solution steps

\int 12 t (3 t^{2} - 1)^{3} d t

Take the constant out:

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

= 12 \cdot \int t (3 t^{2} - 1)^{3} d t

Apply u-substitution

: \int \frac{u ^{3}}{6} d u

= 12 \cdot \int \frac{u ^{3}}{6} d u

Take the constant out:

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

= 12 \cdot \frac{1}{6} \cdot \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= 12 \cdot \frac{1}{6} \cdot \frac{u ^{4}}{4}

Substitute back

u = 3 t^{2} - 1

= 12 \cdot \frac{1}{6} \cdot \frac{( 3 t ^{2} - 1 ) ^{4}}{4}

Simplify

12 \cdot \frac{1}{6} \cdot \frac{( 3 t ^{2} - 1 ) ^{4}}{4} : \frac{1}{2} (3 t^{2} - 1)^{4}

= \frac{1}{2} (3 t^{2} - 1)^{4}

Add a constant to the solution

= \frac{1}{2} (3 t^{2} - 1)^{4} + C

e^{- 3 x} + e^{4 x}

x \to 2 lim (2 x^{2} + 6)

x \to 0 + lim (ln (\frac{2}{x ^{2}}))

\int_{0}^{\frac{π}{2}} 4 sin^{2} (3 x) d x

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

What is the integral of 12t(3t^2-1)^3 ?
The integral of 12t(3t^2-1)^3 is 1/2 (3t^2-1)^4+C