What is the integral of (2t^3-3i)^2(6t^2) ?

The integral of (2t^3-3i)^2(6t^2) is 1/3 (-54t^3+8t^9)-12it^6+C

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integral of (2t^3-3i)^2(6t^2)

\int (2 t^{3} - 3 i)^{2} (6 t^{2}) d t

\frac{1}{3} (- 54 t^{3} + 8 t^{9}) - 12 i t^{6} + C

Solution steps

\int (2 t^{3} - 3 i)^{2} \cdot 6 t^{2} d t

Take the constant out:

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

= 6 \cdot \int (2 t^{3} - 3 i)^{2} t^{2} d t

Expand

(2 t^{3} - 3 i)^{2} t^{2} : (4 t^{8} - 9 t^{2}) - 12 i t^{5}

= 6 \cdot \int 4 t^{8} - 9 t^{2} - 12 i t^{5} d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 6 (\int 4 t^{8} d t - \int 9 t^{2} d t - \int 12 i t^{5} d t)

\int 4 t^{8} d t = \frac{4 t ^{9}}{9}

\int 9 t^{2} d t = 3 t^{3}

\int 12 i t^{5} d t = 2 i t^{6}

= 6 (\frac{4 t ^{9}}{9} - 3 t^{3} - 2 i t^{6})

Simplify

6 (\frac{4 t ^{9}}{9} - 3 t^{3} - 2 i t^{6}) : \frac{1}{3} (- 54 t^{3} + 8 t^{9}) - 12 i t^{6}

= \frac{1}{3} (- 54 t^{3} + 8 t^{9}) - 12 i t^{6}

Add a constant to the solution

= \frac{1}{3} (- 54 t^{3} + 8 t^{9}) - 12 i t^{6} + C

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What is the integral of (2t^3-3i)^2(6t^2) ?
The integral of (2t^3-3i)^2(6t^2) is 1/3 (-54t^3+8t^9)-12it^6+C