What is the integral from 0 to 1 of 1/((2x+1)^3) ?

The integral from 0 to 1 of 1/((2x+1)^3) is 2/9

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integral from 0 to 1 of 1/((2x+1)^3)

\int_{0}^{1} \frac{1}{( 2 x + 1 ) ^{3}} d x

\frac{2}{9}

Decimal Notation

0.22222 \dots

Solution steps

\int_{0}^{1} \frac{1}{( 2 x + 1 ) ^{3}} d x

Apply u-substitution

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

= \int_{1}^{3} \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

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

Apply the Power Rule

: [- \frac{1}{2 u ^{2}}]_{1}^{3}

= \frac{1}{2} [- \frac{1}{2 u ^{2}}]_{1}^{3}

Compute the boundaries:

\frac{4}{9}

= \frac{1}{2} \cdot \frac{4}{9}

Simplify

= \frac{2}{9}

V (r) = (\frac{4}{3}) π r^{3}

\frac{d y}{d t} = 3 t^{2} y + (4 t + 1) e^{t^{3}}

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

\int x^{2} x^{2} - 81 d x

What is the integral from 0 to 1 of 1/((2x+1)^3) ?
The integral from 0 to 1 of 1/((2x+1)^3) is 2/9