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

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

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integral of (3x^2+2)(2x^3+4x+1)^{1/2}

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u ^{\frac{1}{2}}}{2} d u

= \int \frac{u ^{\frac{1}{2}}}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int u^{\frac{1}{2}} d u

Apply the Power Rule

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

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

Substitute back

u = 2 x^{3} + 4 x + 1

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

Simplify

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

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

Add a constant to the solution

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

x^{2} (\frac{d y}{d x}) - 2 x y = 3 y^{4}

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

\frac{d y}{d x} = 0.6 y

\int e^{x^{3}} \cdot x^{2} d x

\frac{d}{d x} (arctan (sinh (x)))

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