What is the integral of (x^3+x)sqrt(x^4+2x^2) ?

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

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integral of (x^3+x)sqrt(x^4+2x^2)

\int (x^{3} + x) x^{4} + 2 x^{2} d x

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

Solution steps

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

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (sin (x z^{2}) - x y^{2})

\int x a e^{- a x} d x

y^{^{'}} + 3 t y = t, y (0) = 1

\int \frac{9}{x ( ln ( x ) ) ^{2}} d x

x \to 0 lim (x (cot (2 x)))

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