What is the integral of (x^2+x^4)/(4x) ?

The integral of (x^2+x^4)/(4x) is 1/4 ((x^4)/4+(x^2)/2)+C

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integral of (x^2+x^4)/(4x)

integral

\frac{x ^{2} + x ^{4}}{4 x}

\frac{1}{4} (\frac{x ^{4}}{4} + \frac{x ^{2}}{2}) + C

Solution steps

\int \frac{x ^{2} + x ^{4}}{4 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 \frac{x ^{2} + x ^{4}}{x} d x

Simplify

\frac{x ^{2} + x ^{4}}{x} : x (x^{2} + 1)

= \frac{1}{4} \cdot \int x (x^{2} + 1) d x

Expand

x (x^{2} + 1) : x^{3} + x

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

= \frac{1}{4} (\int x^{3} d x + \int x d x)

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

\int x d x = \frac{x ^{2}}{2}

= \frac{1}{4} (\frac{x ^{4}}{4} + \frac{x ^{2}}{2})

Add a constant to the solution

= \frac{1}{4} (\frac{x ^{4}}{4} + \frac{x ^{2}}{2}) + C

in t e g r a l - x + 6

in t e g r a l cot^{4} (x)

in t e g r a l \frac{- 2}{k - 2}

in t e g r a l \frac{1}{x ( x + 1 ) ^{\frac{1}{2}}}

in t e g r a l \frac{x ^{2}}{( x - 2 ) ( x ^{2} + 1 )}

What is the integral of (x^2+x^4)/(4x) ?
The integral of (x^2+x^4)/(4x) is 1/4 ((x^4)/4+(x^2)/2)+C