What is the integral of 3xsqrt(2x^2+1) ?

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

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integral of 3xsqrt(2x^2+1)

\int 3 x 2 x^{2} + 1 d x

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

Solution steps

\int 3 x 2 x^{2} + 1 d x

Take the constant out:

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

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

Apply u-substitution

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

= 3 \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

= 3 \cdot \frac{1}{4} \cdot \int u d u

Apply the Power Rule

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

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

Substitute back

u = 2 x^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (4 z e^{x yz})

x \to - 9 - lim (\frac{- 6 x}{x + 9})

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

x = 28 - 7 y^{2}, x = 7 y^{2} - 28

\frac{d y}{d x} = - y^{2}

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