What is the integral of (4x^3-2x)(x^4-x^2-5)^3 ?

The integral of (4x^3-2x)(x^4-x^2-5)^3 is ((x^4-x^2-5)^4)/4+C

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integral of (4x^3-2x)(x^4-x^2-5)^3

\int (4 x^{3} - 2 x) (x^{4} - x^{2} - 5)^{3} d x

\frac{( x ^{4} - x ^{2} - 5 ) ^{4}}{4} + C

Solution steps

\int (4 x^{3} - 2 x) (x^{4} - x^{2} - 5)^{3} d x

Apply u-substitution

: \int u^{3} d u

= \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= \frac{u ^{4}}{4}

Substitute back

u = x^{4} - x^{2} - 5

= \frac{( x ^{4} - x ^{2} - 5 ) ^{4}}{4}

Add a constant to the solution

= \frac{( x ^{4} - x ^{2} - 5 ) ^{4}}{4} + C

x \to + 0 + lim (arctan (\frac{2}{x}))

\frac{d}{d x} \frac{8}{( x ^{3} + x ^{2} + 1 ) ^{3}}

\int sin (a x) d x

y = \frac{2 x - sin ( 2 x )}{x ^{2} + cos ( x ^{2} )}

\int \frac{( 1 0 ^{2 x} + 1 )}{( 1 0 ^{x} - 1 )} d x

What is the integral of (4x^3-2x)(x^4-x^2-5)^3 ?
The integral of (4x^3-2x)(x^4-x^2-5)^3 is ((x^4-x^2-5)^4)/4+C