What is the integral of 6x^3sqrt(3x^4+2) ?

The integral of 6x^3sqrt(3x^4+2) is 1/3 (3x^4+2)^{3/2}+C

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integral of 6x^3sqrt(3x^4+2)

\int 6 x^{3} 3 x^{4} + 2 d x

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

Solution steps

\int 6 x^{3} 3 x^{4} + 2 d x

Take the constant out:

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

= 6 \cdot \int x^{3} 3 x^{4} + 2 d x

Apply u-substitution

: \int \frac{u}{12} d u

= 6 \cdot \int \frac{u}{12} d u

Take the constant out:

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

= 6 \cdot \frac{1}{12} \cdot \int u d u

Apply the Power Rule

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

= 6 \cdot \frac{1}{12} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 3 x^{4} + 2

= 6 \cdot \frac{1}{12} \cdot \frac{2}{3} (3 x^{4} + 2)^{\frac{3}{2}}

Simplify

6 \cdot \frac{1}{12} \cdot \frac{2}{3} (3 x^{4} + 2)^{\frac{3}{2}} : \frac{1}{3} (3 x^{4} + 2)^{\frac{3}{2}}

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

Add a constant to the solution

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

\int x ln (3 + x) d x

(- 9 x^{3} + 5 x^{2})^{4}

\int (4 sec (x) tan (x) + 3 sec^{2} (x)) d x

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

\int u^{- 3} \frac{1}{2} d u

What is the integral of 6x^3sqrt(3x^4+2) ?
The integral of 6x^3sqrt(3x^4+2) is 1/3 (3x^4+2)^{3/2}+C