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

Lösung

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

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

Schritte zur Lösung

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

Wende U-Substitution an

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

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

= \int u - 1 d u

Wende die Summenregel an:

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

= \int u d u - \int 1 d u

\int u d u = \frac{2}{3} u^{\frac{3}{2}}

\int 1 d u = u

= \frac{2}{3} u^{\frac{3}{2}} - u

Setze in

u = x + 1

ein

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

Vereinfache

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

Füge eine Konstante zur Lösung hinzu

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

\int \frac{4 x ^{2} - x + 1}{x ^{3} - x ^{2} + x - 1} d x

\int ln ∣ x - 1 ∣ d x

d er i v a t i v e f (x) = \frac{x}{1 - x ^{2}}

d er i v a t i v e f (x) = \frac{2500 x}{ln ( 8 x + 8 )}

x y^{^{'}} + (x + \frac{1}{3}) y = e^{- 2 x} y^{- 2}