What is the integral of (12x^2)/(sqrt((4x^3+5))) ?

The integral of (12x^2)/(sqrt((4x^3+5))) is 2sqrt(4x^3+5)+C

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integral of (12x^2)/(sqrt((4x^3+5)))

\int \frac{12 x ^{2}}{( 4 x ^{3} + 5 )} d x

2 4 x^{3} + 5 + C

Solution steps

\int \frac{12 x ^{2}}{4 x ^{3} + 5} d x

Take the constant out:

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

= 12 \cdot \int \frac{x ^{2}}{4 x ^{3} + 5} d x

Apply u-substitution

: \int \frac{u}{2 x 4 u ^{\frac{3}{2}} + 5} d u

= 12 \cdot \int \frac{1}{3 4 u + 5} d u

Take the constant out:

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

= 12 \cdot \frac{1}{3} \cdot \int \frac{1}{4 u + 5} d u

Apply u-substitution

: \int \frac{1}{4 v} d v

= 12 \cdot \frac{1}{3} \cdot \int \frac{1}{4 v} d v

Take the constant out:

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

= 12 \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot \int \frac{1}{v} d v

Apply the Power Rule

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

= 12 \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot 2 v^{\frac{1}{2}}

Substitute back

= 12 \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot 2 (4 x^{3} + 5)^{\frac{1}{2}}

Simplify

12 \cdot \frac{1}{3} \cdot \frac{1}{4} \cdot 2 (4 x^{3} + 5)^{\frac{1}{2}} : 2 4 x^{3} + 5

= 2 4 x^{3} + 5

Add a constant to the solution

= 2 4 x^{3} + 5 + C

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What is the integral of (12x^2)/(sqrt((4x^3+5))) ?
The integral of (12x^2)/(sqrt((4x^3+5))) is 2sqrt(4x^3+5)+C