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

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

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

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

\frac{2}{3}

Decimal

0.66666 \dots

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 3 \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}}

= 3 \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

= 3 (\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}

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

Simplify

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

= \frac{2}{3}

\int_{y}^{1} 2 d x

\int_{1}^{3} \frac{3 - 4 x}{x ^{3}} d x

\int_{0}^{x} x d x

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

\int_{0}^{1} x e^{- x} - x^{2} d x

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