integral of 1/(sqrt(x+1)+1)

Lösung

\int \frac{1}{x + 1 + 1} d x

- 2 ln x + 1 + 1 + 2 x + 1 + 2 + C

Schritte zur Lösung

\int \frac{1}{x + 1 + 1} d x

\frac{1}{x + 1 + 1} = - \frac{x + 1}{x + 1 + 1} + 1

= \int - \frac{x + 1}{x + 1 + 1} + 1 d x

Wende die Summenregel an:

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

= - \int \frac{x + 1}{x + 1 + 1} d x + \int 1 d x

\int \frac{x + 1}{x + 1 + 1} d x = x + 2 ln x + 1 + 1 - 2 x + 1 - 2

\int 1 d x = x

= - (x + 2 ln x + 1 + 1 - 2 x + 1 - 2) + x

Vereinfache

= - 2 ln x + 1 + 1 + 2 x + 1 + 2

Füge eine Konstante zur Lösung hinzu

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

\int (7 x + 3)^{4} d x

\int \frac{x ^{2} e ^{- x} + x}{x ^{2}} d x

\int \frac{2 x - 1}{3} d x

\int sin (3 x) cos^{3} (3 x) d x

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