integral of 1/(xsqrt(x)+x^2)

Lösung

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

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

Schritte zur Lösung

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

Wende U-Substitution an

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

Entferne die Konstante:

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

= 2 \cdot \int \frac{1}{u ^{2} ( u + 1 )} d u

Ermittle den Partialbruch von

\frac{1}{u ^{2} ( u + 1 )} : - \frac{1}{u} + \frac{1}{u ^{2}} + \frac{1}{u + 1}

= 2 \cdot \int - \frac{1}{u} + \frac{1}{u ^{2}} + \frac{1}{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

= 2 (- \int \frac{1}{u} d u + \int \frac{1}{u ^{2}} d u + \int \frac{1}{u + 1} d u)

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

\int \frac{1}{u + 1} d u = ln ∣ u + 1 ∣

= 2 (- ln ∣ u ∣ - \frac{1}{u} + ln ∣ u + 1 ∣)

Setze in

u = x

ein

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

Füge eine Konstante zur Lösung hinzu

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

\int_{0}^{L} sin^{2} (\frac{nπ x}{L}) d x

d er i v a t i v e y = \frac{e ^{x}}{1 + x}

d er i v a t i v e tan^{2} (6 x)

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

\frac{\partial}{\partial y} (38 - x^{2} - 4 y^{2})