integral of sqrt(t^2+5)

Lösung

\int t^{2} + 5 d t

5 (\frac{1}{10} t 5 + t^{2} + \frac{1}{2} ln \frac{1}{5} t + \frac{1}{5} (5 + t^{2})) + C

Schritte zur Lösung

\int t^{2} + 5 d t

Trigonometrische Substitution anwenden

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

Entferne die Konstante:

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

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

Wende integrale Reduktion an

= 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) ∣

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

Setze in

u = arctan (\frac{1}{5} t)

ein

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

Vereinfache

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

= 5 (\frac{1}{10} t 5 + t^{2} + \frac{1}{2} ln \frac{1}{5} t + \frac{1}{5} (5 + t^{2}))

Füge eine Konstante zur Lösung hinzu

= 5 (\frac{1}{10} t 5 + t^{2} + \frac{1}{2} ln \frac{1}{5} t + \frac{1}{5} (5 + t^{2})) + C

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

\int \frac{1}{cos ( 5 x )} d x

\int \frac{2}{4 x} d x

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\int t (sec^{2} (t^{2})) (tan^{4} (t^{2})) d t