What is the integral of (2x^4-1)(5x^3+6x) ?

The integral of (2x^4-1)(5x^3+6x) is (5x^8)/4+2x^6-(5x^4)/4-3x^2+C

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integral of (2x^4-1)(5x^3+6x)

\int (2 x^{4} - 1) (5 x^{3} + 6 x) d x

\frac{5 x ^{8}}{4} + 2 x^{6} - \frac{5 x ^{4}}{4} - 3 x^{2} + C

Solution steps

\int (2 x^{4} - 1) (5 x^{3} + 6 x) d x

Expand

(2 x^{4} - 1) (5 x^{3} + 6 x) : 10 x^{7} + 12 x^{5} - 5 x^{3} - 6 x

= \int 10 x^{7} + 12 x^{5} - 5 x^{3} - 6 x 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

= \int 10 x^{7} d x + \int 12 x^{5} d x - \int 5 x^{3} d x - \int 6 x d x

\int 10 x^{7} d x = \frac{5 x ^{8}}{4}

\int 12 x^{5} d x = 2 x^{6}

\int 5 x^{3} d x = \frac{5 x ^{4}}{4}

\int 6 x d x = 3 x^{2}

= \frac{5 x ^{8}}{4} + 2 x^{6} - \frac{5 x ^{4}}{4} - 3 x^{2}

Add a constant to the solution

= \frac{5 x ^{8}}{4} + 2 x^{6} - \frac{5 x ^{4}}{4} - 3 x^{2} + C

f (- 3) = - 2 x^{2} + 4 x - \frac{7}{x ^{2}}

\frac{d y}{d x} = y (x y^{4} - 1)

f (x) = - \frac{4}{x}

n = 0 \sum \infty 5 (\frac{3}{4})^{n}

\int sech^{4} (x) tanh (x) d x

What is the integral of (2x^4-1)(5x^3+6x) ?
The integral of (2x^4-1)(5x^3+6x) is (5x^8)/4+2x^6-(5x^4)/4-3x^2+C