What is the integral of 6/((1-2x)^3) ?

The integral of 6/((1-2x)^3) is 3/(2(-2x+1)^2)+C

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integral of 6/((1-2x)^3)

\int \frac{6}{( 1 - 2 x ) ^{3}} d x

\frac{3}{2 ( - 2 x + 1 ) ^{2}} + C

Solution steps

\int \frac{6}{( 1 - 2 x ) ^{3}} d x

Take the constant out:

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

= 6 \cdot \int \frac{1}{( 1 - 2 x ) ^{3}} d x

Apply u-substitution

: \int - \frac{1}{2 u ^{3}} d u

= 6 \cdot \int - \frac{1}{2 u ^{3}} d u

Take the constant out:

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

= 6 (- \frac{1}{2} \cdot \int \frac{1}{u ^{3}} d u)

Apply the Power Rule

: - \frac{1}{2 u ^{2}}

= 6 (- \frac{1}{2} (- \frac{1}{2 u ^{2}}))

Substitute back

u = 1 - 2 x

= 6 (- \frac{1}{2} (- \frac{1}{2 ( 1 - 2 x ) ^{2}}))

Simplify

6 (- \frac{1}{2} (- \frac{1}{2 ( 1 - 2 x ) ^{2}})) : \frac{3}{2 ( - 2 x + 1 ) ^{2}}

= \frac{3}{2 ( - 2 x + 1 ) ^{2}}

Add a constant to the solution

= \frac{3}{2 ( - 2 x + 1 ) ^{2}} + C

x \frac{d y}{d x} + y = - 6 x y^{2}

x \to 2 lim (2 x^{2} - 4 x)

\int \frac{3 x ^{2}}{1 - 6 x ^{3}} d x

\frac{\partial}{\partial x} (y sin (z))

x \to - 8 - lim (\frac{x + 2}{x + 8})

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