What is the integral of (r^2-10r+4)/(r^3) ?

The integral of (r^2-10r+4)/(r^3) is ln|r|+(10)/r-2/(r^2)+C

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

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

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

Solution steps

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

Expand

\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

Apply the Sum Rule:

\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}}

Simplify

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

Add a constant to the solution

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

y^{^{'}} + y = sin (t), y (0) = - 1

x \to - 6 lim (\frac{x ^{2} - 4}{6 - x})

\int 2 t - 1 d t

4 x + 9

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

What is the integral of (r^2-10r+4)/(r^3) ?
The integral of (r^2-10r+4)/(r^3) is ln|r|+(10)/r-2/(r^2)+C