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

The integral of 4x^2(4x-5)^2 is 4((16x^5)/5-10x^4+(25x^3)/3)+C

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

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

4 (\frac{16 x ^{5}}{5} - 10 x^{4} + \frac{25 x ^{3}}{3}) + C

Solution steps

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

Take the constant out:

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

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

Expand

x^{2} (4 x - 5)^{2} : 16 x^{4} - 40 x^{3} + 25 x^{2}

= 4 \cdot \int 16 x^{4} - 40 x^{3} + 25 x^{2} 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

= 4 (\int 16 x^{4} d x - \int 40 x^{3} d x + \int 25 x^{2} d x)

\int 16 x^{4} d x = \frac{16 x ^{5}}{5}

\int 40 x^{3} d x = 10 x^{4}

\int 25 x^{2} d x = \frac{25 x ^{3}}{3}

= 4 (\frac{16 x ^{5}}{5} - 10 x^{4} + \frac{25 x ^{3}}{3})

Add a constant to the solution

= 4 (\frac{16 x ^{5}}{5} - 10 x^{4} + \frac{25 x ^{3}}{3}) + C

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What is the integral of 4x^2(4x-5)^2 ?
The integral of 4x^2(4x-5)^2 is 4((16x^5)/5-10x^4+(25x^3)/3)+C