What is the integral of 2x^4(2-3x^5)^2 ?

The integral of 2x^4(2-3x^5)^2 is 2((4x^5)/5-(6x^{10})/5+(3x^{15})/5)+C

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integral of 2x^4(2-3x^5)^2

\int 2 x^{4} (2 - 3 x^{5})^{2} d x

2 (\frac{4 x ^{5}}{5} - \frac{6 x ^{10}}{5} + \frac{3 x ^{15}}{5}) + C

Solution steps

\int 2 x^{4} (2 - 3 x^{5})^{2} d x

Take the constant out:

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

= 2 \cdot \int x^{4} (2 - 3 x^{5})^{2} d x

Expand

x^{4} (2 - 3 x^{5})^{2} : 4 x^{4} - 12 x^{9} + 9 x^{14}

= 2 \cdot \int 4 x^{4} - 12 x^{9} + 9 x^{14} d x

Apply the Sum Rule:

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

= 2 (\int 4 x^{4} d x - \int 12 x^{9} d x + \int 9 x^{14} d x)

\int 4 x^{4} d x = \frac{4 x ^{5}}{5}

\int 12 x^{9} d x = \frac{6 x ^{10}}{5}

\int 9 x^{14} d x = \frac{3 x ^{15}}{5}

= 2 (\frac{4 x ^{5}}{5} - \frac{6 x ^{10}}{5} + \frac{3 x ^{15}}{5})

Add a constant to the solution

= 2 (\frac{4 x ^{5}}{5} - \frac{6 x ^{10}}{5} + \frac{3 x ^{15}}{5}) + C

What is the integral of 2x^4(2-3x^5)^2 ?
The integral of 2x^4(2-3x^5)^2 is 2((4x^5)/5-(6x^{10})/5+(3x^{15})/5)+C