integral from 0 to x of 3/((1+x)^4)

Lösung

\int_{0}^{x} \frac{3}{( 1 + x ) ^{4}} d x

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

Schritte zur Lösung

\int_{0}^{x} \frac{3}{( 1 + x ) ^{4}} d x

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int_{0}^{x} \frac{1}{( 1 + x ) ^{4}} d x

Wende U-Substitution an

= 3 \cdot \int_{1}^{1 + x} \frac{1}{u ^{4}} d u

Wende die Potenzregel an

= 3 [- \frac{1}{3 u ^{3}}]_{1}^{1 + x}

Berechne die Grenzen:

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

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

Vereinfache

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

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