integral of (5x^2)/(x+7)

Lösung

\int \frac{5 x ^{2}}{x + 7} d x

5 (\frac{( x + 7 ) ^{2}}{2} - 14 (x + 7) + 49 ln ∣ x + 7 ∣) + C

Schritte zur Lösung

\int \frac{5 x ^{2}}{x + 7} d x

Entferne die Konstante:

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

= 5 \cdot \int \frac{x ^{2}}{x + 7} d x

Wende U-Substitution an

= 5 \cdot \int \frac{( u - 7 ) ^{2}}{u} d u

Multipliziere aus

\frac{( u - 7 ) ^{2}}{u} : u - 14 + \frac{49}{u}

= 5 \cdot \int u - 14 + \frac{49}{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

= 5 (\int u d u - \int 14 d u + \int \frac{49}{u} d u)

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

\int 14 d u = 14 u

\int \frac{49}{u} d u = 49 ln ∣ u ∣

= 5 (\frac{u ^{2}}{2} - 14 u + 49 ln ∣ u ∣)

Setze in

u = x + 7

ein

= 5 (\frac{( x + 7 ) ^{2}}{2} - 14 (x + 7) + 49 ln ∣ x + 7 ∣)

Füge eine Konstante zur Lösung hinzu

= 5 (\frac{( x + 7 ) ^{2}}{2} - 14 (x + 7) + 49 ln ∣ x + 7 ∣) + C

x \to 3 - lim (\frac{7}{x - 3})

x \to 0 lim (lo g_{e} (x^{2}))

x \to 1 lim (x - 1 + 4)

l a pl a ce t r an s f or m t \cdot cos (3 t)

\frac{d}{d x} (x^{5} e^{4 x})