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

The integral of x(2x^2-1)^4 is -1/20 (-2x^2+1)^5+C

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

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

- \frac{1}{20} (- 2 x^{2} + 1)^{5} + C

Solution steps

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

Factor

2 x^{2} - 1 : - (- 2 x^{2} + 1)

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

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

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

Apply u-substitution

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

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

Apply the Power Rule

: \frac{u ^{5}}{5}

= - \frac{1}{4} \cdot \frac{u ^{5}}{5}

Substitute back

u = - 2 x^{2} + 1

= - \frac{1}{4} \cdot \frac{( - 2 x ^{2} + 1 ) ^{5}}{5}

Simplify

- \frac{1}{4} \cdot \frac{( - 2 x ^{2} + 1 ) ^{5}}{5} : - \frac{1}{20} (- 2 x^{2} + 1)^{5}

= - \frac{1}{20} (- 2 x^{2} + 1)^{5}

Add a constant to the solution

= - \frac{1}{20} (- 2 x^{2} + 1)^{5} + C

\frac{d}{d x} (2 x + y)

\frac{d}{d t} (t^{2} e^{- t})

x \cdot y^{2} \cdot y^{^{'}} = x^{3} + y^{3}

\int \frac{1}{1 + ( 2 x + 5 ) ^{2}} d x

11 x - 6 y x^{2} + 1 \frac{d y}{d x} = 0

What is the integral of x(2x^2-1)^4 ?
The integral of x(2x^2-1)^4 is -1/20 (-2x^2+1)^5+C