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

Lösung

\int_{2}^{3} \frac{x ^{2}}{x ^{2} - 1} d x

- \frac{ln ( 2 ) - ln ( 3 )}{2} + 1

Dezimale

1.20273 \dots

Schritte zur Lösung

\int_{2}^{3} \frac{x ^{2}}{x ^{2} - 1} d x

Faktorisiere

x^{2} - 1 : - (- x^{2} + 1)

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

Entferne die Konstante:

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

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

\frac{x ^{2}}{- x ^{2} + 1} = \frac{1}{- x ^{2} + 1} - 1

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

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{2}^{3} \frac{1}{- x ^{2} + 1} d x = \frac{ln ( 2 ) - ln ( 3 )}{2}

\int_{2}^{3} 1 d x = 1

= - (\frac{ln ( 2 ) - ln ( 3 )}{2} - 1)

Vereinfache

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

\int_{0}^{4} x^{2} - 9 d x

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

\int_{4}^{5} 16 - x^{2} d x

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

\int_{0}^{1} - x^{2} + 2 d x