integral from 1 to b of 1/((x+8)^2)

Lösung

\int_{1}^{b} \frac{1}{( x + 8 ) ^{2}} d x

- \frac{1}{b + 8} + \frac{1}{9}

Schritte zur Lösung

\int_{1}^{b} \frac{1}{( x + 8 ) ^{2}} d x

Wende U-Substitution an

= \int_{9}^{b + 8} \frac{1}{u ^{2}} d u

Wende die Potenzregel an

= [- \frac{1}{u}]_{9}^{b + 8}

Berechne die Grenzen:

- \frac{1}{b + 8} + \frac{1}{9}

= - \frac{1}{b + 8} + \frac{1}{9}