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

Lösung

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

\frac{1}{15} (3 x^{2} - 2 x + 2) 2 x + 1 + C

Schritte zur Lösung

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

Wende U-Substitution an

= \int \frac{( u ^{2} - 1 ) ^{2}}{4} d u

Entferne die Konstante:

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

= \frac{1}{4} \cdot \int (u^{2} - 1)^{2} d u

Multipliziere aus

(u^{2} - 1)^{2} : u^{4} - 2 u^{2} + 1

= \frac{1}{4} \cdot \int u^{4} - 2 u^{2} + 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

= \frac{1}{4} (\int u^{4} d u - \int 2 u^{2} d u + \int 1 d u)

\int u^{4} d u = \frac{u ^{5}}{5}

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

\int 1 d u = u

= \frac{1}{4} (\frac{u ^{5}}{5} - \frac{2 u ^{3}}{3} + u)

Setze in

u = 2 x + 1

ein

= \frac{1}{4} (\frac{( 2 x + 1 ) ^{5}}{5} - \frac{2 ( 2 x + 1 ) ^{3}}{3} + 2 x + 1)

Vereinfache

\frac{1}{4} (\frac{( 2 x + 1 ) ^{5}}{5} - \frac{2 ( 2 x + 1 ) ^{3}}{3} + 2 x + 1) : \frac{1}{15} (3 x^{2} - 2 x + 2) 2 x + 1

= \frac{1}{15} (3 x^{2} - 2 x + 2) 2 x + 1

Füge eine Konstante zur Lösung hinzu

= \frac{1}{15} (3 x^{2} - 2 x + 2) 2 x + 1 + C

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

\frac{d}{d x} (5 (2 x cos (7 x) - 7 x^{2} sin (7 x)))

\int cos^{3} (20 x) d x

y^{^{''}} + 2 y = sin (2 x), y (0) = 0, y^{^{'}} (0) = 0

x \to 2 + lim (\frac{∣ x ∣}{2 - x})