What is the integral of 1/((3-5x)^4) ?

The integral of 1/((3-5x)^4) is 1/(15(3-5x)^3)+C

integral of 1/((3-5x)^4)

\int \frac{1}{( 3 - 5 x ) ^{4}} d x

\frac{1}{15 ( 3 - 5 x ) ^{3}} + C

Solution steps

\int \frac{1}{( 3 - 5 x ) ^{4}} d x

Apply u-substitution

= \int - \frac{1}{5 u ^{4}} d u

Take the constant out:

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

= - \frac{1}{5} \cdot \int \frac{1}{u ^{4}} d u

Apply the Power Rule

= - \frac{1}{5} (- \frac{1}{3 u ^{3}})

Substitute back

u = 3 - 5 x

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

Simplify

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

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

Add a constant to the solution

= \frac{1}{15 ( 3 - 5 x ) ^{3}} + C

\int 12 - 4 x d x

d er i v a t i v e 6 x - 4

\int (tan (x) - cot (x))^{2} d x

s l o p e (x - 3)^{2} + (y - 1)^{2} = 0

d er i v a t i v e y = arcsin (4 x + 1)

What is the integral of 1/((3-5x)^4) ?
The integral of 1/((3-5x)^4) is 1/(15(3-5x)^3)+C