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

The integral of x(2x^2-5) is 1/8 (2x^2-5)^2+C

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integral of x(2x^2-5)

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

\frac{1}{8} (2 x^{2} - 5)^{2} + C

Solution steps

\int x (2 x^{2} - 5) 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{u ^{2}}{2}

= \frac{1}{4} \cdot \frac{u ^{2}}{2}

Substitute back

u = 2 x^{2} - 5

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

Simplify

\frac{1}{4} \cdot \frac{( 2 x ^{2} - 5 ) ^{2}}{2} : \frac{1}{8} (2 x^{2} - 5)^{2}

= \frac{1}{8} (2 x^{2} - 5)^{2}

Add a constant to the solution

= \frac{1}{8} (2 x^{2} - 5)^{2} + C

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What is the integral of x(2x^2-5) ?
The integral of x(2x^2-5) is 1/8 (2x^2-5)^2+C