What is the integral of 4/((2x+1)^3) ?

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

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integral of 4/((2x+1)^3)

\int \frac{4}{( 2 x + 1 ) ^{3}} d x

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

Solution steps

\int \frac{4}{( 2 x + 1 ) ^{3}} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{( 2 x + 1 ) ^{3}} d x

Apply u-substitution

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

= 4 \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

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{u ^{3}} d u

Apply the Power Rule

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

= 4 \cdot \frac{1}{2} (- \frac{1}{2 u ^{2}})

Substitute back

u = 2 x + 1

= 4 \cdot \frac{1}{2} (- \frac{1}{2 ( 2 x + 1 ) ^{2}})

Simplify

4 \cdot \frac{1}{2} (- \frac{1}{2 ( 2 x + 1 ) ^{2}}) : - \frac{1}{( 2 x + 1 ) ^{2}}

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

Add a constant to the solution

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

\frac{d}{d x} ((x + 1) (x + 2))

x \to 0 lim (x - 8)

\frac{d P}{d t} = - 0.02 P

\int_{0}^{\infty} x e^{2 x} d x

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

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