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

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

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integral of (x^2+1/4)x

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

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

Solution steps

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

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

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

Simplify

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

Expand

x (4 x^{2} + 1) : 4 x^{3} + x

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

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

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

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

Add a constant to the solution

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

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