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

The integral from 0 to 1 of x(1+2x)^{-3} is 1/18

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integral from 0 to 1 of x(1+2x)^{-3}

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

\frac{1}{18}

Decimal

0.05555 \dots

Solution steps

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

Simplify

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

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

Apply u-substitution

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

= \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

= \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}}

= \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

= \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}

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

Simplify

\frac{1}{4} (\frac{2}{3} - \frac{4}{9}) : \frac{1}{18}

= \frac{1}{18}

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What is the integral from 0 to 1 of x(1+2x)^{-3} ?
The integral from 0 to 1 of x(1+2x)^{-3} is 1/18