integral of sqrt(9+25x^2)

Lösung

\int 9 + 25 x^{2} d x

\frac{1}{10} (5 x 25 x^{2} + 9 + 9 ln (\frac{5 x + 25 x ^{2} + 9}{3})) + C

Schritte zur Lösung

\int 9 + 25 x^{2} d x

Trigonometrische Substitution anwenden

= \int \frac{9}{5} sec^{3} (u) d u

Entferne die Konstante:

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

= \frac{9}{5} \cdot \int sec^{3} (u) d u

Wende integrale Reduktion an

= \frac{9}{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{9}{5} (\frac{sec ^{2} ( u ) sin ( u )}{2} + \frac{1}{2} ln ∣ tan (u) + sec (u) ∣)

Setze in

u = arctan (\frac{5}{3} x)

ein

= \frac{9}{5} (\frac{sec ^{2} ( arctan ( \frac{5}{3} x ) ) sin ( arctan ( \frac{5}{3} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{5}{3} x)) + sec (arctan (\frac{5}{3} x)))

Vereinfache

\frac{9}{5} (\frac{sec ^{2} ( arctan ( \frac{5}{3} x ) ) sin ( arctan ( \frac{5}{3} x ) )}{2} + \frac{1}{2} ln tan (arctan (\frac{5}{3} x)) + sec (arctan (\frac{5}{3} x))) : \frac{1}{10} (5 x 25 x^{2} + 9 + 9 ln (\frac{5 x + 25 x ^{2} + 9}{3}))

= \frac{1}{10} (5 x 25 x^{2} + 9 + 9 ln (\frac{5 x + 25 x ^{2} + 9}{3}))

Füge eine Konstante zur Lösung hinzu

= \frac{1}{10} (5 x 25 x^{2} + 9 + 9 ln (\frac{5 x + 25 x ^{2} + 9}{3})) + C

x \to 4 lim ((\frac{x ^{4}}{4}) + 1)

l a pl a ce t r an s f or m t cos (t)

\int (x^{2} - 1) d x

y^{^{''}} - 10 y^{^{'}} + 9 y = 27

\frac{d}{d x} (x^{2} - 4 x + 4)