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

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

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

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

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

Solution steps

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

Simplify

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

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

x^{2} = (x),

assuming

x \geq 0

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

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

Simplify

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

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

Add a constant to the solution

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

\int_{0}^{3} 2 x - x^{2} d x

x \to 5 lim (\frac{x + 3}{x - 3})

\int sin (5 x) sin (3 x) d x

\int \frac{48 x ^{2}}{x ^{4} - 74 x ^{2} + 1225} d x

- \frac{1}{s}

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