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

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

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

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

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

Solution steps

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

Take the constant out:

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

= 3 \cdot \int \frac{1}{( 5 v - 1 ) ^{4}} d v

Apply u-substitution

: \int \frac{1}{5 u ^{4}} d u

= 3 \cdot \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

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

Apply the Power Rule

: - \frac{1}{3 u ^{3}}

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

Substitute back

u = 5 v - 1

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

Simplify

3 \cdot \frac{1}{5} (- \frac{1}{3 ( 5 v - 1 ) ^{3}}) : - \frac{1}{5 ( 5 v - 1 ) ^{3}}

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

Add a constant to the solution

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

\frac{d}{d x} (cos (x) - x)

(x - 5)^{2} + (y - 3)^{2} = 18, (8, 6)

\int_{0}^{4} π (3 x)^{2} d x

f (x) = cos (tan (x))

y^{^{''}} + a y^{^{'}} + b y = t + e^{3 t}

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