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Beliebt Rechnen >

integral of (7x^2)/(sqrt(4x^2-5))

Lösung

\int \frac{7 x ^{2}}{4 x ^{2} - 5} d x

Lösung

\frac{7}{16} (2 x 4 x^{2} - 5 + 5 (ln 2 x + 5 \frac{4 x ^{2} - 5}{5} - \frac{ln ( 5 )}{2})) + C

Schritte zur Lösung

\int \frac{7 x ^{2}}{4 x ^{2} - 5} d x

Entferne die Konstante:

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

= 7 \cdot \int \frac{x ^{2}}{4 x ^{2} - 5} d x

Trigonometrische Substitution anwenden

= 7 \cdot \int \frac{5 sec ^{3} ( u )}{8} d u

Entferne die Konstante:

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

= 7 \cdot \frac{5}{8} \cdot \int sec^{3} (u) d u

Wende integrale Reduktion an

= 7 \cdot \frac{5}{8} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} \cdot \int sec (u) d u)

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= 7 \cdot \frac{5}{8} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Setze in

u = arcsec (\frac{2}{5} x)

ein

= 7 \cdot \frac{5}{8} \frac{sec ^{2} ( arcsec ( \frac{2}{5} x ) ) sin ( arcsec ( \frac{2}{5} x ) )}{2} + \frac{1}{2} ln tan (arcsec (\frac{2}{5} x)) + sec (arcsec (\frac{2}{5} x))

Vereinfache

7 \cdot \frac{5}{8} \frac{sec ^{2} ( arcsec ( \frac{2}{5} x ) ) sin ( arcsec ( \frac{2}{5} x ) )}{2} + \frac{1}{2} ln tan (arcsec (\frac{2}{5} x)) + sec (arcsec (\frac{2}{5} x)) : \frac{7}{16} (2 x 4 x^{2} - 5 + 5 (ln 2 x + 5 \frac{4 x ^{2} - 5}{5} - \frac{ln ( 5 )}{2}))

= \frac{7}{16} (2 x 4 x^{2} - 5 + 5 (ln 2 x + 5 \frac{4 x ^{2} - 5}{5} - \frac{ln ( 5 )}{2}))

Füge eine Konstante zur Lösung hinzu

= \frac{7}{16} (2 x 4 x^{2} - 5 + 5 (ln 2 x + 5 \frac{4 x ^{2} - 5}{5} - \frac{ln ( 5 )}{2})) + C

Graph

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