What is the integral of y^3(4y^2-25) ?

The integral of y^3(4y^2-25) is (2y^6)/3-(25y^4)/4+C

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integral of y^3(4y^2-25)

\int y^{3} (4 y^{2} - 25) d y

\frac{2 y ^{6}}{3} - \frac{25 y ^{4}}{4} + C

Solution steps

\int y^{3} (4 y^{2} - 25) d y

Expand

y^{3} (4 y^{2} - 25) : 4 y^{5} - 25 y^{3}

= \int 4 y^{5} - 25 y^{3} d y

Apply the Sum Rule:

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

= \int 4 y^{5} d y - \int 25 y^{3} d y

\int 4 y^{5} d y = \frac{2 y ^{6}}{3}

\int 25 y^{3} d y = \frac{25 y ^{4}}{4}

= \frac{2 y ^{6}}{3} - \frac{25 y ^{4}}{4}

Add a constant to the solution

= \frac{2 y ^{6}}{3} - \frac{25 y ^{4}}{4} + C

n = 4 \sum \infty \frac{1}{n}

\int 8 x^{3} (2 x^{4} + 4)^{4} d x

1 + (\frac{2}{5})^{\circ} x (\frac{2}{3})^{\circ} - 25

g (x) = [ln (5 x - 2)]^{4}

\frac{1}{12 x ^{3}}

What is the integral of y^3(4y^2-25) ?
The integral of y^3(4y^2-25) is (2y^6)/3-(25y^4)/4+C