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

The integral of sqrt(x^2+2x^4) is 1/6 (1+2x^2)^{3/2}+C

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

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

\frac{1}{6} (1 + 2 x^{2})^{\frac{3}{2}} + C

Solution steps

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

Factor

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

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

x^{2} = (x),

assuming

x \geq 0

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

Apply u-substitution

: \int \frac{u}{4} d u

= \int \frac{u}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int u d u

Apply the Power Rule

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

= \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 1 + 2 x^{2}

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

Simplify

\frac{1}{4} \cdot \frac{2}{3} (1 + 2 x^{2})^{\frac{3}{2}} : \frac{1}{6} (1 + 2 x^{2})^{\frac{3}{2}}

= \frac{1}{6} (1 + 2 x^{2})^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{6} (1 + 2 x^{2})^{\frac{3}{2}} + C

\frac{d}{d x} (\frac{sin ^{5} ( x )}{5} - \frac{sin ^{7} ( x )}{7})

\frac{d}{d x} (e^{x} - cos (x))

y = (2 x + 1) (x + 5)^{3}

n = 0 \sum \infty (- 1)^{n} (2^{n})

\frac{d}{d x} (- \frac{1}{3 x ^{3}})

What is the integral of sqrt(x^2+2x^4) ?
The integral of sqrt(x^2+2x^4) is 1/6 (1+2x^2)^{3/2}+C