integral from 1 to 2 of 1/((2x+3y)^2)

Lösung

\int_{1}^{2} \frac{1}{( 2 x + 3 y ) ^{2}} d y

\frac{1}{2 ( x + 3 ) ( 2 x + 3 )}

Schritte zur Lösung

\int_{1}^{2} \frac{1}{( 2 x + 3 y ) ^{2}} d y

Wende U-Substitution an

= \int_{2 x + 3}^{2 x + 6} \frac{1}{3 u ^{2}} d u

Entferne die Konstante:

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

= \frac{1}{3} \cdot \int_{2 x + 3}^{2 x + 6} \frac{1}{u ^{2}} d u

Wende die Potenzregel an

= \frac{1}{3} [- \frac{1}{u}]_{2 x + 3}^{2 x + 6}

Berechne die Grenzen:

- \frac{1}{2 x + 6} + \frac{1}{2 x + 3}

= \frac{1}{3} (- \frac{1}{2 x + 6} + \frac{1}{2 x + 3})

Vereinfache

= \frac{1}{2 ( x + 3 ) ( 2 x + 3 )}

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