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

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

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

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

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

Solution steps

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

Simplify

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

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

Apply u-substitution

: \int u^{2} d u

= \int u^{2} d u

Apply the Power Rule

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

= \frac{u ^{3}}{3}

Substitute back

u = 4 + x^{2}

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

Simplify

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

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

Add a constant to the solution

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

t d t + y e^{- t} d y = 0

\frac{d}{d x} (5 x + 2)

\int \frac{3 x ^{2} + 1}{x ^{2} \cdot 4 - x ^{2}} d x

\frac{\partial}{\partial x} (sin (x \cdot y \cdot e^{x^{2}}))

f^{^{'}} (x) = 2^{4 x} + 4 x^{2}

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