What is the integral of x(x^2+1)sqrt(4-2x^2-x^4) ?

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

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integral of x(x^2+1)sqrt(4-2x^2-x^4)

\int x (x^{2} + 1) 4 - 2 x^{2} - x^{4} d x

- \frac{1}{6} (4 - 2 x^{2} - x^{4})^{\frac{3}{2}} + C

Solution steps

\int x (x^{2} + 1) 4 - 2 x^{2} - x^{4} 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 = 4 - 2 x^{2} - x^{4}

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

Simplify

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

= - \frac{1}{6} (4 - 2 x^{2} - x^{4})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{6} (4 - 2 x^{2} - x^{4})^{\frac{3}{2}} + C

\int 16 + 25 x^{2} d x

x \to 2 lim (\frac{x - 3}{x - 2})

\int 2 t - 5 d t

\frac{d y}{d x} = \frac{x}{y}

\frac{d y}{d x} = \frac{( 2 sec ( y ) )}{( x - 4 ) ^{2}}

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