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

The integral of sqrt(4x^4+x^2) is 1/12 (4x^2+1)^{3/2}+C

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

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

\frac{1}{12} (4 x^{2} + 1)^{\frac{3}{2}} + C

Solution steps

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

Factor

4 x^{4} + x^{2} : x^{2} 4 x^{2} + 1

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

x^{2} = (x),

assuming

x \geq 0

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

Apply u-substitution

: \int \frac{u}{8} d u

= \int \frac{u}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{8} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 4 x^{2} + 1

= \frac{1}{8} \cdot \frac{2}{3} (4 x^{2} + 1)^{\frac{3}{2}}

Simplify

\frac{1}{8} \cdot \frac{2}{3} (4 x^{2} + 1)^{\frac{3}{2}} : \frac{1}{12} (4 x^{2} + 1)^{\frac{3}{2}}

= \frac{1}{12} (4 x^{2} + 1)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{12} (4 x^{2} + 1)^{\frac{3}{2}} + C

f (x) = \frac{4}{( 3 x ) ^{- 2}}

\int \frac{5}{x ^{- 1} + 5} d x

x \to \infty lim (\frac{1}{x} - 3)

\frac{d}{d x} (\frac{7 x ^{2}}{x})

\int_{0}^{1} 9 (x^{2} + 1) e^{- x} d x

What is the integral of sqrt(4x^4+x^2) ?
The integral of sqrt(4x^4+x^2) is 1/12 (4x^2+1)^{3/2}+C