integral of sqrt(4+25x^2)

Lösung

\int 4 + 25 x^{2} d x

\frac{1}{10} (5 x 25 x^{2} + 4 + 4 ln (\frac{5 x + 25 x ^{2} + 4}{2})) + C

Schritte zur Lösung

\int 4 + 25 x^{2} d x

Trigonometrische Substitution anwenden

= \int \frac{4}{5} sec^{3} (u) d u

Entferne die Konstante:

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

= \frac{4}{5} \cdot \int sec^{3} (u) d u

Wende integrale Reduktion an

= \frac{4}{5} (\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{4}{5} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Setze in

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

ein

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

Vereinfache

\frac{4}{5} (\frac{sec ^{2} ( arctan ( \frac{5}{2} x ) ) sin ( arctan ( \frac{5}{2} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{5}{2} x)) + sec (arctan (\frac{5}{2} x))) : \frac{1}{10} (5 x 25 x^{2} + 4 + 4 ln (\frac{5 x + 25 x ^{2} + 4}{2}))

= \frac{1}{10} (5 x 25 x^{2} + 4 + 4 ln (\frac{5 x + 25 x ^{2} + 4}{2}))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{10} (5 x 25 x^{2} + 4 + 4 ln (\frac{5 x + 25 x ^{2} + 4}{2})) + C

\frac{d}{d x} (sin (3 x + y z))

\frac{d}{d x} (x^{0.7})

d er i v a t i v e y = x^{2} - 2 x - 24 x

4 y^{^{''}} + 9 y^{^{'}} = 0

y^{^{''}} + y = 6 sin (2 t) + 4 t cos (2 t)