integral from 0 to 1 of (4x)/((x^2+1)^2)

Lösung

\int_{0}^{1} \frac{4 x}{( x ^{2} + 1 ) ^{2}} d x

1

Schritte zur Lösung

\int_{0}^{1} \frac{4 x}{( x ^{2} + 1 ) ^{2}} d x

Entferne die Konstante:

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

= 4 \cdot \int_{0}^{1} \frac{x}{( x ^{2} + 1 ) ^{2}} d x

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Wende die Potenzregel an

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

Vereinfache

4 \cdot \frac{1}{2} [- \frac{1}{u}]_{1}^{2} : 2 [- \frac{1}{u}]_{1}^{2}

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

Berechne die Grenzen:

\frac{1}{2}

= 2 \cdot \frac{1}{2}

Vereinfache

= 1

x \to 5 lim (\frac{3 x + 5}{5 x + 6})

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

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7 \frac{d y}{d x} = y^{2}