What is the integral of (4x^2)/(sqrt(7x^3+6)) ?

The integral of (4x^2)/(sqrt(7x^3+6)) is 8/21 sqrt(7x^3+6)+C

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integral of (4x^2)/(sqrt(7x^3+6))

\int \frac{4 x ^{2}}{7 x ^{3} + 6} d x

\frac{8}{21} 7 x^{3} + 6 + C

Solution steps

\int \frac{4 x ^{2}}{7 x ^{3} + 6} d x

Take the constant out:

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

= 4 \cdot \int \frac{x ^{2}}{7 x ^{3} + 6} d x

Apply u-substitution

: \int \frac{u}{2 x 7 u ^{\frac{3}{2}} + 6} d u

= 4 \cdot \int \frac{1}{3 7 u + 6} d u

Take the constant out:

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

= 4 \cdot \frac{1}{3} \cdot \int \frac{1}{7 u + 6} d u

Apply u-substitution

: \int \frac{1}{7 v} d v

= 4 \cdot \frac{1}{3} \cdot \int \frac{1}{7 v} d v

Take the constant out:

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

= 4 \cdot \frac{1}{3} \cdot \frac{1}{7} \cdot \int \frac{1}{v} d v

Apply the Power Rule

: 2 v^{\frac{1}{2}}

= 4 \cdot \frac{1}{3} \cdot \frac{1}{7} \cdot 2 v^{\frac{1}{2}}

Substitute back

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

Simplify

4 \cdot \frac{1}{3} \cdot \frac{1}{7} \cdot 2 (7 x^{3} + 6)^{\frac{1}{2}} : \frac{8}{21} 7 x^{3} + 6

= \frac{8}{21} 7 x^{3} + 6

Add a constant to the solution

= \frac{8}{21} 7 x^{3} + 6 + C

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What is the integral of (4x^2)/(sqrt(7x^3+6)) ?
The integral of (4x^2)/(sqrt(7x^3+6)) is 8/21 sqrt(7x^3+6)+C