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

The integral from 0 to 1 of x(x+1)^{-3} is 1/8

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

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

\frac{1}{8}

Decimal Notation

0.125

Solution steps

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

Simplify

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

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

Apply u-substitution

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

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

Expand

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

= \int_{1}^{2} \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

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

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

\int_{1}^{2} \frac{1}{u ^{3}} d u = \frac{3}{8}

= \frac{1}{2} - \frac{3}{8}

Simplify

\frac{1}{2} - \frac{3}{8} : \frac{1}{8}

= \frac{1}{8}

\int_{0}^{1} 1 + u^{2} d u

\int_{0}^{π} 2 sin (x) - sin (2 x) d x

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

\int_{π}^{2 π} cos (θ) d θ

\int_{- 2}^{2} 8 - x^{2} d x

What is the integral from 0 to 1 of x(x+1)^{-3} ?
The integral from 0 to 1 of x(x+1)^{-3} is 1/8