integral from 0 to x of 1/(t^2-25)

Lösung

\int_{0}^{x} \frac{1}{t ^{2} - 25} d t

- \frac{ln \frac{x}{5} + 1 - ln \frac{x}{5} - 1}{10}

Schritte zur Lösung

\int_{0}^{x} \frac{1}{t ^{2} - 25} d t

Wende integrale Substitution an

= \int_{0}^{\frac{x}{5}} \frac{1}{5 ( u ^{2} - 1 )} d u

Entferne die Konstante:

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

= \frac{1}{5} \cdot \int_{0}^{\frac{x}{5}} \frac{1}{u ^{2} - 1} d u

Faktorisiere

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

= \frac{1}{5} \cdot \int_{0}^{\frac{x}{5}} \frac{1}{- ( - u ^{2} + 1 )} d u

Entferne die Konstante:

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

= \frac{1}{5} (- \int_{0}^{\frac{x}{5}} \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{1}{5} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{\frac{x}{5}})

Vereinfache

\frac{1}{5} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{0}^{\frac{x}{5}}) : - \frac{1}{5} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{0}^{\frac{x}{5}}

= - \frac{1}{5} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{0}^{\frac{x}{5}}

Berechne die Grenzen:

\frac{ln \frac{x}{5} + 1 - ln \frac{x}{5} - 1}{2}

= - \frac{1}{5} \cdot \frac{ln \frac{x}{5} + 1 - ln \frac{x}{5} - 1}{2}

Vereinfache

= - \frac{ln \frac{x}{5} + 1 - ln \frac{x}{5} - 1}{10}

y^{^{'}} = y + 3 t e^{t}, y (0) = 4

x \to 3 lim (x - 8)

\int (s^{3} + 3)^{2} (1 - s^{2})^{3} d s

y^{^{''}} - 16 y = 2 e^{4 x}

\int_{1}^{2} t ln (t) d t