integral from 0 to 9 of 1/((1+x)^2)

Lösung

\int_{0}^{9} \frac{1}{( 1 + x ) ^{2}} d x

\frac{9}{10}

Dezimale

0.9

Schritte zur Lösung

\int_{0}^{9} \frac{1}{( 1 + x ) ^{2}} d x

Wende U-Substitution an

= \int_{1}^{10} \frac{1}{u ^{2}} d u

Wende die Potenzregel an

= [- \frac{1}{u}]_{1}^{10}

Berechne die Grenzen:

\frac{9}{10}

= \frac{9}{10}

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