What is the integral of (x^4)/4*1/x ?

The integral of (x^4)/4*1/x is (x^4)/(16)+C

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integral of (x^4)/4*1/x

\int \frac{x ^{4}}{4} \cdot \frac{1}{x} d x

\frac{x ^{4}}{16} + C

Solution steps

\int \frac{x ^{4}}{4} \cdot \frac{1}{x} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int x^{4} \frac{1}{x} d x

Simplify

x^{4} \frac{1}{x} : x^{3}

= \frac{1}{4} \cdot \int x^{3} d x

Apply the Power Rule

: \frac{x ^{4}}{4}

= \frac{1}{4} \cdot \frac{x ^{4}}{4}

Simplify

\frac{1}{4} \cdot \frac{x ^{4}}{4} : \frac{x ^{4}}{16}

= \frac{x ^{4}}{16}

Add a constant to the solution

= \frac{x ^{4}}{16} + C

y^{^{'}} = t^{2} e^{- 4 t} - 4 y

f (x) = 4 x (x^{2} - 4 x + 5)^{7}, at x = 2

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

f (x) = e^{4 x}

\frac{s}{s ^{2} + 6 s + 25}

What is the integral of (x^4)/4*1/x ?
The integral of (x^4)/4*1/x is (x^4)/(16)+C