integral of 1/(sqrt(2x-1)-\sqrt[4]{2x-1)}

Lösung

\int \frac{1}{2 x - 1 - 4 2 x - 1} d x

2 ln 4 2 x - 1 - 1 + 2 x - 1 + 2 4 2 x - 1 - 3 + C

Schritte zur Lösung

\int \frac{1}{2 x - 1 - 4 2 x - 1} d x

Wende U-Substitution an

= \int \frac{u}{u - 1} d u

Wende U-Substitution an

= \int \frac{v ^{2} + 2 v + 1}{0.5 v} d v

Entferne die Konstante:

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

= \frac{1}{0.5} \cdot \int \frac{v ^{2} + 2 v + 1}{v} d v

Vereinfache

= 2 \cdot \int \frac{v ^{2} + 2 v + 1}{v} d v

Multipliziere aus

\frac{v ^{2} + 2 v + 1}{v} : v + 2 + \frac{1}{v}

= 2 \cdot \int v + 2 + \frac{1}{v} d v

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int v d v + \int 2 d v + \int \frac{1}{v} d v)

\int v d v = \frac{v ^{2}}{2}

\int 2 d v = 2 v

\int \frac{1}{v} d v = ln ∣ v ∣

= 2 (\frac{v ^{2}}{2} + 2 v + ln ∣ v ∣)

Ersetze zurück

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

Vereinfache

2 \frac{( 2 x - 1 - 1 ) ^{2}}{2} + 2 (2 x - 1 - 1) + ln 2 x - 1 - 1 : 2 ln 4 2 x - 1 - 1 + 2 x - 1 + 2 4 2 x - 1 - 3

= 2 ln 4 2 x - 1 - 1 + 2 x - 1 + 2 4 2 x - 1 - 3

Füge eine Konstante zur Lösung hinzu

= 2 ln 4 2 x - 1 - 1 + 2 x - 1 + 2 4 2 x - 1 - 3 + C

\int x^{1.1} (\frac{1}{3 x} - 1) d x

im pl i c i t x^{2} + y^{2} = 25

\int_{1}^{2} x (x - 1)^{9} d x

h \to 1 - lim (\frac{h - 2}{1 - h})

\int tan (6 x) sec^{2} (6 x) d x