What is the integral of 2x^2sqrt(4x^3-7) ?

The integral of 2x^2sqrt(4x^3-7) is 1/9 (4x^3-7)^{3/2}+C

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integral of 2x^2sqrt(4x^3-7)

\int 2 x^{2} 4 x^{3} - 7 d x

\frac{1}{9} (4 x^{3} - 7)^{\frac{3}{2}} + C

Solution steps

\int 2 x^{2} 4 x^{3} - 7 d x

Take the constant out:

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

= 2 \cdot \int x^{2} 4 x^{3} - 7 d x

Apply u-substitution

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

= 2 \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

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

Apply the Power Rule

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

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

Substitute back

u = 4 x^{3} - 7

= 2 \cdot \frac{1}{6} \cdot \frac{( 4 x ^{3} - 7 ) ^{3}}{3}

Simplify

2 \cdot \frac{1}{6} \cdot \frac{( 4 x ^{3} - 7 ) ^{3}}{3} : \frac{1}{9} (4 x^{3} - 7)^{\frac{3}{2}}

= \frac{1}{9} (4 x^{3} - 7)^{\frac{3}{2}}

Add a constant to the solution

= \frac{1}{9} (4 x^{3} - 7)^{\frac{3}{2}} + C

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What is the integral of 2x^2sqrt(4x^3-7) ?
The integral of 2x^2sqrt(4x^3-7) is 1/9 (4x^3-7)^{3/2}+C