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

Lösung

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

\frac{1}{8}

Dezimale

0.125

Schritte zur Lösung

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

Vereinfache

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

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

Wende U-Substitution an

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

Multipliziere aus

\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

Wende die Summenregel an:

\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}

Vereinfache

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

= \frac{1}{8}

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