What is the integral of 6x^3sqrt(3x^4+8) ?

The integral of 6x^3sqrt(3x^4+8) is 1/3 (3x^4+8)^{3/2}+C

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integral of 6x^3sqrt(3x^4+8)

\int 6 x^{3} 3 x^{4} + 8 d x

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

Solution steps

\int 6 x^{3} 3 x^{4} + 8 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} + 8 d x

Apply u-substitution

: \int \frac{u ^{2}}{6} d u

= 6 \cdot \int \frac{u ^{2}}{6} d u

Take the constant out:

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

= 6 \cdot \frac{1}{6} \cdot \int u^{2} d u

Apply the Power Rule

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

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

Substitute back

u = 3 x^{4} + 8

= 6 \cdot \frac{1}{6} \cdot \frac{( 3 x ^{4} + 8 ) ^{3}}{3}

Simplify

6 \cdot \frac{1}{6} \cdot \frac{( 3 x ^{4} + 8 ) ^{3}}{3} : \frac{1}{3} (3 x^{4} + 8)^{\frac{3}{2}}

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

Add a constant to the solution

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

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What is the integral of 6x^3sqrt(3x^4+8) ?
The integral of 6x^3sqrt(3x^4+8) is 1/3 (3x^4+8)^{3/2}+C