integral of [ 1/(sqrt(x))+sin(x-1)]

Lösung

\int [\frac{1}{x} + sin (x - 1)] d x

2 x + x sin (x - 1) + x sin (1 - x) - cos (1 - x) + C

Schritte zur Lösung

\int \frac{1}{x} + sin (x - 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{1}{x} d x + \int sin (x - 1) d x

\int \frac{1}{x} d x = 2 x

\int sin (x - 1) d x = x sin (x - 1) + x sin (1 - x) - cos (1 - x)

= 2 x + x sin (x - 1) + x sin (1 - x) - cos (1 - x)

Füge eine Konstante zur Lösung hinzu

= 2 x + x sin (x - 1) + x sin (1 - x) - cos (1 - x) + C

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

\int \frac{1}{sin ( \frac{y}{2} ) cos ( \frac{y}{2} )} d y

\int \frac{sin ( 2 x )}{1 - cos ( 2 x )} d x

\int sin (2 x) e^{5 x} d x

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