integral of (r^2-10r+4)/(r^3)

Lösung

\int \frac{r ^{2} - 10 r + 4}{r ^{3}} d r

ln ∣ r ∣ + \frac{10}{r} - \frac{2}{r ^{2}} + C

Schritte zur Lösung

\int \frac{r ^{2} - 10 r + 4}{r ^{3}} d r

Multipliziere aus

\frac{r ^{2} - 10 r + 4}{r ^{3}} : \frac{1}{r} - \frac{10}{r ^{2}} + \frac{4}{r ^{3}}

= \int \frac{1}{r} - \frac{10}{r ^{2}} + \frac{4}{r ^{3}} d r

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{r} d r - \int \frac{10}{r ^{2}} d r + \int \frac{4}{r ^{3}} d r

\int \frac{1}{r} d r = ln ∣ r ∣

\int \frac{10}{r ^{2}} d r = - \frac{10}{r}

\int \frac{4}{r ^{3}} d r = - \frac{2}{r ^{2}}

= ln ∣ r ∣ - (- \frac{10}{r}) - \frac{2}{r ^{2}}

Vereinfache

= ln ∣ r ∣ + \frac{10}{r} - \frac{2}{r ^{2}}

Füge eine Konstante zur Lösung hinzu

= ln ∣ r ∣ + \frac{10}{r} - \frac{2}{r ^{2}} + C

\int (4 x - 3)^{5} d x

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y^{^{''}} + 9 y = 3 sin (3 t)

\frac{d}{d x} (ln (16))

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