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

Lösung

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

- \frac{1}{1 + x} + 1

Schritte zur Lösung

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

Wende U-Substitution an

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

Wende die Potenzregel an

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

Berechne die Grenzen:

- \frac{1}{1 + x} + 1

= - \frac{1}{1 + x} + 1

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