integral of xln(x^2+3)

Lösung

\int x ln (x^{2} + 3) d x

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

Schritte zur Lösung

\int x ln (x^{2} + 3) d x

Wende U-Substitution an

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

Entferne die Konstante:

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

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

Wende die partielle Integration an

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

\int 1 d u = u

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

Setze in

u = x^{2} + 3

ein

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

Multipliziere aus

(x^{2} + 3) ln (x^{2} + 3) - (x^{2} + 3) : x^{2} ln (x^{2} + 3) + 3 ln (x^{2} + 3) - x^{2} - 3

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

Füge eine Konstante zur Lösung hinzu

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

\int_{0}^{\frac{π}{39}} tan (13 x) d x

\int \frac{x ^{3} - x}{x} d x

\frac{\partial}{\partial x} (x^{4} y^{9} + 6 x^{7} y)

y^{^{'}} - \frac{7}{x} \cdot y = 0

\int \frac{1}{1 - cos ( x ) + sin ( x )} d x