What is the integral of x^4sqrt(3x^5-5) ?

The integral of x^4sqrt(3x^5-5) is 2/45 (3x^5-5)^{3/2}+C

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integral of x^4sqrt(3x^5-5)

\int x^{4} 3 x^{5} - 5 d x

\frac{2}{45} (3 x^{5} - 5)^{\frac{3}{2}} + C

Solution steps

\int x^{4} 3 x^{5} - 5 d x

Apply u-substitution

: \int \frac{u}{15} d u

= \int \frac{u}{15} d u

Take the constant out:

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

= \frac{1}{15} \cdot \int u d u

Apply the Power Rule

: \frac{2}{3} u^{\frac{3}{2}}

= \frac{1}{15} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 3 x^{5} - 5

= \frac{1}{15} \cdot \frac{2}{3} (3 x^{5} - 5)^{\frac{3}{2}}

Simplify

\frac{1}{15} \cdot \frac{2}{3} (3 x^{5} - 5)^{\frac{3}{2}} : \frac{2}{45} (3 x^{5} - 5)^{\frac{3}{2}}

= \frac{2}{45} (3 x^{5} - 5)^{\frac{3}{2}}

Add a constant to the solution

= \frac{2}{45} (3 x^{5} - 5)^{\frac{3}{2}} + C

\frac{d}{d x} (\frac{5}{x x})

(z, 2), (- 2, - 6)

\int 6 x sec (3 x^{2}) tan (3 x^{2}) d x

x^{2} y^{^{''}} + x y^{^{'}} + 25 y = 0

\frac{\partial}{\partial x} (x^{2} + y^{2})

What is the integral of x^4sqrt(3x^5-5) ?
The integral of x^4sqrt(3x^5-5) is 2/45 (3x^5-5)^{3/2}+C