integral of (x^2)/(x+15)

Lösung

\int \frac{x ^{2}}{x + 15} d x

\frac{( x + 15 ) ^{2}}{2} - 30 (x + 15) + 225 ln ∣ x + 15 ∣ + C

Schritte zur Lösung

\int \frac{x ^{2}}{x + 15} d x

Wende U-Substitution an

= \int \frac{( u - 15 ) ^{2}}{u} d u

Multipliziere aus

\frac{( u - 15 ) ^{2}}{u} : u - 30 + \frac{225}{u}

= \int u - 30 + \frac{225}{u} 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

= \int u d u - \int 30 d u + \int \frac{225}{u} d u

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

\int 30 d u = 30 u

\int \frac{225}{u} d u = 225 ln ∣ u ∣

= \frac{u ^{2}}{2} - 30 u + 225 ln ∣ u ∣

Setze in

u = x + 15

ein

= \frac{( x + 15 ) ^{2}}{2} - 30 (x + 15) + 225 ln ∣ x + 15 ∣

Füge eine Konstante zur Lösung hinzu

= \frac{( x + 15 ) ^{2}}{2} - 30 (x + 15) + 225 ln ∣ x + 15 ∣ + C

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

\int (- 4 x^{6} - 5 x^{5} - 2 x^{2} + 5 - e^{x}) d x

y^{^{''}} - 2 y^{^{'}} = x + 2 e^{x}, y (0) = 17, y^{^{'}} (0) = \frac{55}{4}

\frac{\partial}{\partial y} (ln (1 + x^{2} + e^{y} + z))

in v erse l a pl a ce \frac{s ^{2} + 6 s + 2}{s ( s + 1 ) ^{2}}