integral from 0 to 0.6 of 1/((1-x)(3-x))

Lösung

\int_{0}^{0.6} \frac{1}{( 1 - x ) ( 3 - x )} d x

0.34657 \dots

Schritte zur Lösung

\int_{0}^{0.6} \frac{1}{( 1 - x ) ( 3 - x )} d x

Vervollständige das Quadrat

(1 - x) (3 - x) : (x - 2)^{2} - 1

= \int_{0}^{0.6} \frac{1}{( x - 2 ) ^{2} - 1} d x

Wende U-Substitution an

= \int_{- 2}^{- 1.4} \frac{1}{u ^{2} - 1} d u

Faktorisiere

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

= \int_{- 2}^{- 1.4} \frac{1}{- ( - u ^{2} + 1 )} d u

Entferne die Konstante:

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

= - \int_{- 2}^{- 1.4} \frac{1}{- u ^{2} + 1} d u

Nutze das gemeinsame Integral :

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

= - [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{- 2}^{- 1.4}

Vereinfache

- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{- 2}^{- 1.4} : - [\frac{1}{2} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣)]_{- 2}^{- 1.4}

= - [\frac{1}{2} (ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣)]_{- 2}^{- 1.4}

Berechne die Grenzen:

- 0.34657 \dots

= - (- 0.34657 \dots)

Vereinfache

= 0.34657 \dots

\int \frac{1}{t ^{2} 2 - t ^{2}} d t

d er i v a t i v e \frac{( cos ( x ) )}{cos ^{2} ( \cdot x ) - 1}

\int (n + 3) 3 d n

\int \frac{2}{e ^{x} + 2} d x

t an g e n t f (x) = \frac{36 x}{x ^{2} + 36}, at x = - 3