What is the integral of (xsqrt(x^4)) ?

The integral of (xsqrt(x^4)) is (x^4)/4+C

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integral of (xsqrt(x^4))

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

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

Solution steps

\int x x^{4} d x

x^{4} = x^{2},

assuming

x \geq 0

= \int x x^{2} d x

Simplify

x x^{2} : x^{3}

= \int x^{3} d x

Apply the Power Rule

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{1}{2} ln (\frac{1 + x}{1 - x}))

y^{^{'''}} - 5 y^{^{''}} + 8 y^{^{'}} + 14 y = 0

\int \frac{12 x ^{3} + 6 x ^{2}}{3 x ^{4} + 2 x ^{3} - 5} d x

\frac{\partial}{\partial t} (1 + t)

What is the integral of (xsqrt(x^4)) ?
The integral of (xsqrt(x^4)) is (x^4)/4+C