integral of 1/(sqrt((5t+1)^2-1))

Lösung

\int \frac{1}{( 5 t + 1 ) ^{2} - 1} d t

\frac{1}{5} ln 25 t^{2} + 10 t + 5 t + 1 + C

Schritte zur Lösung

\int \frac{1}{( 5 t + 1 ) ^{2} - 1} d t

Multipliziere aus

(5 t + 1)^{2} - 1 : 25 t^{2} + 10 t

= \int \frac{1}{25 t ^{2} + 10 t} d t

Vervollständige das Quadrat

25 t^{2} + 10 t : 25 (t + \frac{1}{5})^{2} - 1

= \int \frac{1}{25 ( t + \frac{1}{5} ) ^{2} - 1} d t

Wende U-Substitution an

= \int \frac{1}{25 u ^{2} - 1} d u

Trigonometrische Substitution anwenden

= \int \frac{sec ( v )}{5} d v

Entferne die Konstante:

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

= \frac{1}{5} \cdot \int sec (v) d v

Nutze das gemeinsame Integral :

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

= \frac{1}{5} ln ∣ tan (v) + sec (v) ∣

Ersetze zurück

= \frac{1}{5} ln tan (arcsec (5 (t + \frac{1}{5}))) + sec (arcsec (5 (t + \frac{1}{5})))

Vereinfache

\frac{1}{5} ln tan (arcsec (5 (t + \frac{1}{5}))) + sec (arcsec (5 (t + \frac{1}{5}))) : \frac{1}{5} ln 25 t^{2} + 10 t + 5 t + 1

= \frac{1}{5} ln 25 t^{2} + 10 t + 5 t + 1

Füge eine Konstante zur Lösung hinzu

= \frac{1}{5} ln 25 t^{2} + 10 t + 5 t + 1 + C

\int \frac{1 - ln ( x )}{x ln ( x )} d x

\int \frac{8 x - 24}{( x ^{2} + 2 x + 4 ) ^{2}} d x

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

\int \frac{4 x ^{3} - 1}{x ^{4} - x + 4} d x

\int x (x - 4)^{9} d x