What is the integral of t(2t^2+4)^3 ?

The integral of t(2t^2+4)^3 is 1/16 (2t^2+4)^4+C

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integral of t(2t^2+4)^3

\int t (2 t^{2} + 4)^{3} d t

\frac{1}{16} (2 t^{2} + 4)^{4} + C

Solution steps

\int t (2 t^{2} + 4)^{3} d t

Apply u-substitution

: \int \frac{u ^{3}}{4} d u

= \int \frac{u ^{3}}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int u^{3} d u

Apply the Power Rule

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

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

Substitute back

u = 2 t^{2} + 4

= \frac{1}{4} \cdot \frac{( 2 t ^{2} + 4 ) ^{4}}{4}

Simplify

\frac{1}{4} \cdot \frac{( 2 t ^{2} + 4 ) ^{4}}{4} : \frac{1}{16} (2 t^{2} + 4)^{4}

= \frac{1}{16} (2 t^{2} + 4)^{4}

Add a constant to the solution

= \frac{1}{16} (2 t^{2} + 4)^{4} + C

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What is the integral of t(2t^2+4)^3 ?
The integral of t(2t^2+4)^3 is 1/16 (2t^2+4)^4+C