integral from-1 to 0 of 1/(4x^2-9)

Lösung

\int_{- 1}^{0} \frac{1}{4 x ^{2} - 9} d x

- \frac{ln ( 5 )}{12}

Dezimale

- 0.13411 \dots

Schritte zur Lösung

\int_{- 1}^{0} \frac{1}{4 x ^{2} - 9} d x

Wende integrale Substitution an

= \int_{- \frac{2}{3}}^{0} \frac{1}{6 ( u ^{2} - 1 )} d u

Entferne die Konstante:

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

= \frac{1}{6} \cdot \int_{- \frac{2}{3}}^{0} \frac{1}{u ^{2} - 1} d u

Faktorisiere

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

= \frac{1}{6} \cdot \int_{- \frac{2}{3}}^{0} \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}{6} (- \int_{- \frac{2}{3}}^{0} \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}{6} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{- \frac{2}{3}}^{0})

Vereinfache

\frac{1}{6} (- [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{- \frac{2}{3}}^{0}) : - \frac{1}{6} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{- \frac{2}{3}}^{0}

= - \frac{1}{6} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{- \frac{2}{3}}^{0}

Berechne die Grenzen:

- \frac{- ln ( 3 ) - ln ( \frac{5}{3} )}{2}

= - \frac{1}{6} (- \frac{- ln ( 3 ) - ln ( \frac{5}{3} )}{2})

Vereinfache

= - \frac{ln ( 5 )}{12}

\int_{1}^{e^{x}} 4 ln (t) d t

\int_{- 3}^{5} (4 x - 3)^{3} d x

\int_{0}^{\frac{π}{2}} \frac{1}{cos ( x )} d x

\int_{- 2}^{- 1} \frac{1}{( 11 + 5 x ) ^{3}} d x

\int_{- 1}^{5} (x^{2} - 4 x - 5) d x