integral of ln(1+r^2)r

Lösung

\int ln (1 + r^{2}) r d r

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

Schritte zur Lösung

\int ln (1 + r^{2}) r d r

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 = 1 + r^{2}

ein

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

Multipliziere aus

(1 + r^{2}) ln (1 + r^{2}) - (1 + r^{2}) : ln (1 + r^{2}) + r^{2} ln (1 + r^{2}) - 1 - r^{2}

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

Füge eine Konstante zur Lösung hinzu

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

\frac{\partial}{\partial x} (yz ln (x y))

y (x + y + 1) d x + (x + 2 y) d y = 0

x^{^{''}} - 2 x - 5 = 0, x (0) = 1, x (1) = 3

\int \frac{39}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

x \to 2 lim (2 x - 3)