What is the integral of sqrt(1+2x^4)(8x^3) ?

The integral of sqrt(1+2x^4)(8x^3) is 2/3 (1+2x^4)^{3/2}+C

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integral of sqrt(1+2x^4)(8x^3)

\int 1 + 2 x^{4} (8 x^{3}) d x

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

Solution steps

\int 1 + 2 x^{4} \cdot 8 x^{3} d x

Take the constant out:

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

= 8 \cdot \int 1 + 2 x^{4} x^{3} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= 8 \cdot \frac{1}{8} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 1 + 2 x^{4}

= 8 \cdot \frac{1}{8} \cdot \frac{2}{3} (1 + 2 x^{4})^{\frac{3}{2}}

Simplify

8 \cdot \frac{1}{8} \cdot \frac{2}{3} (1 + 2 x^{4})^{\frac{3}{2}} : \frac{2}{3} (1 + 2 x^{4})^{\frac{3}{2}}

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

Add a constant to the solution

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

y^{^{'}} = \frac{x}{y}

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

\frac{\partial}{\partial x} (x ln (z^{4} + x^{7}))

f (x) = \frac{2 x}{x ^{2} + 1}, at x = - 1

\int e^{4 t} d t

What is the integral of sqrt(1+2x^4)(8x^3) ?
The integral of sqrt(1+2x^4)(8x^3) is 2/3 (1+2x^4)^{3/2}+C