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

The integral of x^4(y-2) is (x^5(-2+y))/5+C

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integral of x^4(y-2)

\int x^{4} (y - 2)

\frac{x ^{5} ( - 2 + y )}{5} + C

Solution steps

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

Take the constant out:

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

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

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

= (- 2 + y) \frac{x ^{5}}{5}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{x ^{5} ( - 2 + y )}{5}

Add a constant to the solution

= \frac{x ^{5} ( - 2 + y )}{5} + C

y = \frac{4 x ^{2} + 6 x + 8}{x}

x^{2} \frac{d y}{d x} - 2 x y = 5 y^{4}, y (1) = \frac{1}{3}

\int (\frac{sec ( x ) + cos ( x )}{2 cos ( x )}) d x

\frac{d}{d x} ((2 x + 9)^{2})

y = 3 + lo g_{2} (x)