integral of (xln(x^2+4))/(x^2+4)

Lösung

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

\frac{1}{4} ln^{2} (x^{2} + 4) + C

Schritte zur Lösung

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

Wende U-Substitution an

= \int \frac{ln ( u + 4 )}{2 ( u + 4 )} d u

Entferne die Konstante:

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

= \frac{1}{2} \cdot \int \frac{ln ( u + 4 )}{u + 4} d u

Wende U-Substitution an

= \frac{1}{2} \cdot \int \frac{ln ( v )}{v} d v

Wende U-Substitution an

= \frac{1}{2} \cdot \int w d w

Wende die Potenzregel an

= \frac{1}{2} \cdot \frac{w ^{2}}{2}

Ersetze zurück

= \frac{1}{2} \cdot \frac{ln ^{2} ( x ^{2} + 4 )}{2}

Vereinfache

\frac{1}{2} \cdot \frac{ln ^{2} ( x ^{2} + 4 )}{2} : \frac{1}{4} ln^{2} (x^{2} + 4)

= \frac{1}{4} ln^{2} (x^{2} + 4)

Füge eine Konstante zur Lösung hinzu

= \frac{1}{4} ln^{2} (x^{2} + 4) + C

\int x^{2} x^{3} + 37 d x

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

t an g e n t f (x) = 7 x + 2, at x = 2

(21 + x^{2}) y^{^{'}} - 2 x y = 0

y^{^{'}} = 9.8 - \frac{3 y}{3.5}