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

The integral of (x^2)/(sqrt(4x^3+5)) is 1/6 sqrt(4x^3+5)+C

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

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

\frac{1}{6} 4 x^{3} + 5 + C

Solution steps

\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

= \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

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

Apply u-substitution

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

= \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

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

Apply the Power Rule

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

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

Substitute back

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

Simplify

\frac{1}{3} \cdot \frac{1}{4} \cdot 2 (4 x^{3} + 5)^{\frac{1}{2}} : \frac{1}{6} 4 x^{3} + 5

= \frac{1}{6} 4 x^{3} + 5

Add a constant to the solution

= \frac{1}{6} 4 x^{3} + 5 + C

f (t) = 1 + cos (t)

\int_{- \infty}^{8} r e^{\frac{r}{4}} d r

\int (1 + e^{y}) d y

\int cos^{4} (x) sin (x) d x

\frac{d y}{d x} = x y + 7 x + 3 y + 21

What is the integral of (x^2)/(sqrt(4x^3+5)) ?
The integral of (x^2)/(sqrt(4x^3+5)) is 1/6 sqrt(4x^3+5)+C