integral of sqrt(5+4x^2)

Lösung

\int 5 + 4 x^{2} d x

\frac{1}{4} (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 5 + 4 x^{2} d x

Trigonometrische Substitution anwenden

= \int \frac{5}{2} sec^{3} (u) d u

Entferne die Konstante:

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

= \frac{5}{2} \cdot \int sec^{3} (u) d u

Wende integrale Reduktion an

= \frac{5}{2} (\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) ∣

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

Setze in

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

ein

= \frac{5}{2} \frac{sec ^{2} ( arctan ( \frac{2}{5} x ) ) sin ( arctan ( \frac{2}{5} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{5} x)) + sec (arctan (\frac{2}{5} x))

Vereinfache

\frac{5}{2} \frac{sec ^{2} ( arctan ( \frac{2}{5} x ) ) sin ( arctan ( \frac{2}{5} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{2}{5} x)) + sec (arctan (\frac{2}{5} x)) : \frac{1}{4} (2 x 4 x^{2} + 5 + 5 (ln 2 x + 5 \frac{4 x ^{2} + 5}{5} - \frac{ln ( 5 )}{2}))

= \frac{1}{4} (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{1}{4} (2 x 4 x^{2} + 5 + 5 (ln 2 x + 5 \frac{4 x ^{2} + 5}{5} - \frac{ln ( 5 )}{2})) + C

d er i v a t i v e ln (2^{x} - 4)

\frac{\partial}{\partial y} (3 cos (x) - sin (x) y)

a re a y = (25 - 4 x^{2})^{2}, x = - \frac{5}{2}, x = \frac{5}{2}

\int \frac{5 x + 3}{x ^{2} + 1} d x

n = 1 \sum \infty 1 9^{n} x^{n} n!