integral from 1 to 3 of 7/(16-x^2)

Lösung

\int_{1}^{3} \frac{7}{16 - x ^{2}} d x

\frac{7 ( ln ( 7 ) - ln ( \frac{5}{4} ) + ln ( \frac{3}{4} ) )}{8}

Dezimale

1.25569 \dots

Schritte zur Lösung

\int_{1}^{3} \frac{7}{16 - x ^{2}} d x

Entferne die Konstante:

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

= 7 \cdot \int_{1}^{3} \frac{1}{16 - x ^{2}} d x

Wende integrale Substitution an

= 7 \cdot \int_{\frac{1}{4}}^{\frac{3}{4}} \frac{1}{4 ( - u ^{2} + 1 )} d u

Entferne die Konstante:

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

= 7 \cdot \frac{1}{4} \cdot \int_{\frac{1}{4}}^{\frac{3}{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}

= 7 \cdot \frac{1}{4} [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{\frac{1}{4}}^{\frac{3}{4}}

Vereinfache

7 \cdot \frac{1}{4} [\frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}]_{\frac{1}{4}}^{\frac{3}{4}} : \frac{7}{4} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{\frac{1}{4}}^{\frac{3}{4}}

= \frac{7}{4} [\frac{ln ∣ u + 1 ∣ - ln ∣ u - 1 ∣}{2}]_{\frac{1}{4}}^{\frac{3}{4}}

Berechne die Grenzen:

\frac{ln ( \frac{7}{4} ) + 2 ln ( 2 ) - ln ( \frac{5}{4} ) + ln ( \frac{3}{4} )}{2}

= \frac{7}{4} \cdot \frac{ln ( \frac{7}{4} ) + 2 ln ( 2 ) - ln ( \frac{5}{4} ) + ln ( \frac{3}{4} )}{2}

Vereinfache

= \frac{7 ( ln ( 7 ) - ln ( \frac{5}{4} ) + ln ( \frac{3}{4} ) )}{8}

\int \frac{26 x ^{3} - 52 x ^{2} + 7}{x ^{2} - 2 x} d x

d er i v a t i v e ln (9)

\int_{0}^{1} \frac{x}{( x ^{2} + 1 ) ^{3}} d x

\int_{- 2}^{0} π (2 x^{2} + 4 x + 3)^{2} d x

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