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

Lösung

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

\frac{b ^{2}}{2 ( 2 b ^{2} + 1 )}

Schritte zur Lösung

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

Wende U-Substitution an

= \int_{1}^{2 b^{2} + 1} \frac{1}{4 u ^{2}} d u

Entferne die Konstante:

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

= \frac{1}{4} \cdot \int_{1}^{2 b^{2} + 1} \frac{1}{u ^{2}} d u

Wende die Potenzregel an

= \frac{1}{4} [- \frac{1}{u}]_{1}^{2 b^{2} + 1}

Berechne die Grenzen:

- \frac{1}{2 b ^{2} + 1} + 1

= \frac{1}{4} (- \frac{1}{2 b ^{2} + 1} + 1)

Vereinfache

= \frac{b ^{2}}{2 ( 2 b ^{2} + 1 )}

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