What is the integral from 0 to 1 of (6x)/((2x+1)^3) ?

The integral from 0 to 1 of (6x)/((2x+1)^3) is 1/3

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integral from 0 to 1 of (6x)/((2x+1)^3)

\int_{0}^{1} \frac{6 x}{( 2 x + 1 ) ^{3}} d x

\frac{1}{3}

Decimal Notation

0.33333 \dots

Solution steps

\int_{0}^{1} \frac{6 x}{( 2 x + 1 ) ^{3}} d x

Take the constant out:

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

= 6 \cdot \int_{0}^{1} \frac{x}{( 2 x + 1 ) ^{3}} d x

Apply u-substitution

: \int_{1}^{3} \frac{u - 1}{4 u ^{3}} d u

= 6 \cdot \int_{1}^{3} \frac{u - 1}{4 u ^{3}} d u

Take the constant out:

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

= 6 \cdot \frac{1}{4} \cdot \int_{1}^{3} \frac{u - 1}{u ^{3}} d u

Expand

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

= 6 \cdot \frac{1}{4} \cdot \int_{1}^{3} \frac{1}{u ^{2}} - \frac{1}{u ^{3}} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 6 \cdot \frac{1}{4} (\int_{1}^{3} \frac{1}{u ^{2}} d u - \int_{1}^{3} \frac{1}{u ^{3}} d u)

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

\int_{1}^{3} \frac{1}{u ^{3}} d u = \frac{4}{9}

= 6 \cdot \frac{1}{4} (\frac{2}{3} - \frac{4}{9})

Simplify

6 \cdot \frac{1}{4} (\frac{2}{3} - \frac{4}{9}) : \frac{1}{3}

= \frac{1}{3}

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What is the integral from 0 to 1 of (6x)/((2x+1)^3) ?
The integral from 0 to 1 of (6x)/((2x+1)^3) is 1/3