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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 4 \cdot \int \frac{u}{4} d u

Take the constant out:

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

= 4 \cdot \frac{1}{4} \cdot \int u d u

Apply the Power Rule

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

= 4 \cdot \frac{1}{4} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 1 + 2 x^{2}

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{x ^{2} - 5}{x} d x

\frac{\partial}{\partial x} (2 y^{2})

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

\int tan^{6} (x) sec^{4} (x) d x

\frac{d}{d x} (4^{(x^{9})})

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