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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

= 2 \cdot \int - \frac{1}{10} d u

Integral of a constant:

\int a d x = a x

= 2 (- \frac{1}{10}) u

Substitute back

u = 3 - 4 x^{5}

= 2 (- \frac{1}{10}) 3 - 4 x^{5}

Simplify

2 (- \frac{1}{10}) 3 - 4 x^{5} : - \frac{1}{5} 3 - 4 x^{5}

= - \frac{1}{5} 3 - 4 x^{5}

Add a constant to the solution

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

t an g e n t f (x) = x^{3} - 5 x^{2} - 2 x + 1, at x = - 2

d er i v a t i v e \frac{3 - sec ( x )}{tan ( x )}

\int_{1}^{2} x ln (8 x) d x

\int \frac{x ^{2} + 1}{( x - 5 ) ( x - 4 ) ^{2}} d x

\frac{d}{d t} (a t cos (4 t) + b t sin (4 t))

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