integral of (x+3)/(sqrt(2x+3))

Lösung

\int \frac{x + 3}{2 x + 3} d x

\frac{1}{3} x 2 x + 3 + 2 2 x + 3 + C

Schritte zur Lösung

\int \frac{x + 3}{2 x + 3} d x

Multipliziere aus

\frac{x + 3}{2 x + 3} : \frac{x}{2 x + 3} + \frac{3}{2 x + 3}

= \int \frac{x}{2 x + 3} + \frac{3}{2 x + 3} 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}{2 x + 3} d x + \int \frac{3}{2 x + 3} d x

\int \frac{x}{2 x + 3} d x = \frac{1}{3} (x - 3) 2 x + 3

\int \frac{3}{2 x + 3} d x = 3 2 x + 3

= \frac{1}{3} (x - 3) 2 x + 3 + 3 2 x + 3

Vereinfache

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

= \frac{1}{3} x 2 x + 3 + 2 2 x + 3

Füge eine Konstante zur Lösung hinzu

= \frac{1}{3} x 2 x + 3 + 2 2 x + 3 + C

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

\int \frac{3 x + 23}{x ^{2} - 8 x - 9} d x

\int (x - π) sin (n x) d x

\int \frac{2}{1 - t ^{2}} d t

\int θ sin^{2} (θ) d θ