What is the integral of x(4x^2+8) ?

The integral of x(4x^2+8) is 1/16 (4x^2+8)^2+C

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integral of x(4x^2+8)

\int x (4 x^{2} + 8) d x

\frac{1}{16} (4 x^{2} + 8)^{2} + C

Solution steps

\int x (4 x^{2} + 8) d x

Apply u-substitution

: \int \frac{u}{8} d u

= \int \frac{u}{8} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{8} \cdot \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= \frac{1}{8} \cdot \frac{u ^{2}}{2}

Substitute back

u = 4 x^{2} + 8

= \frac{1}{8} \cdot \frac{( 4 x ^{2} + 8 ) ^{2}}{2}

Simplify

\frac{1}{8} \cdot \frac{( 4 x ^{2} + 8 ) ^{2}}{2} : \frac{1}{16} (4 x^{2} + 8)^{2}

= \frac{1}{16} (4 x^{2} + 8)^{2}

Add a constant to the solution

= \frac{1}{16} (4 x^{2} + 8)^{2} + C

\int x 1 - x^{4} d x

f (x) = 2 x, at x = 8

\int_{0}^{1} 2 (x^{2} + 1) e^{- x} d x

\int 4 cos (8 x) d x

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

What is the integral of x(4x^2+8) ?
The integral of x(4x^2+8) is 1/16 (4x^2+8)^2+C