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

Lösung

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

\frac{1}{2 ( y + 1 ) ( 2 y + 1 )}

Schritte zur Lösung

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

Wende Exponentenregel an:

a^{- b} = \frac{1}{a ^{b}}

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

Wende U-Substitution an

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

Wende die Potenzregel an

= [- \frac{1}{u}]_{1 + 2 y}^{2 + 2 y}

Berechne die Grenzen:

- \frac{1}{2 + 2 y} + \frac{1}{1 + 2 y}

= - \frac{1}{2 + 2 y} + \frac{1}{1 + 2 y}

Vereinfache

= \frac{1}{2 ( y + 1 ) ( 2 y + 1 )}

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