What is the integral of (r^2+1)/(r-3) ?

The integral of (r^2+1)/(r-3) is 1/2 (r-3)^2+6r-18+10ln|r-3|+C

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integral of (r^2+1)/(r-3)

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

\frac{1}{2} (r - 3)^{2} + 6 r - 18 + 10 ln ∣ r - 3 ∣ + C

Solution steps

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

Expand

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

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

\int \frac{r ^{2}}{r - 3} d r = \frac{( r - 3 ) ^{2}}{2} + 6 (r - 3) + 9 ln ∣ r - 3 ∣

\int \frac{1}{r - 3} d r = ln ∣ r - 3 ∣

= \frac{( r - 3 ) ^{2}}{2} + 6 (r - 3) + 9 ln ∣ r - 3 ∣ + ln ∣ r - 3 ∣

Simplify

\frac{( r - 3 ) ^{2}}{2} + 6 (r - 3) + 9 ln ∣ r - 3 ∣ + ln ∣ r - 3 ∣ : \frac{1}{2} (r - 3)^{2} + 6 r - 18 + 10 ln ∣ r - 3 ∣

= \frac{1}{2} (r - 3)^{2} + 6 r - 18 + 10 ln ∣ r - 3 ∣

Add a constant to the solution

= \frac{1}{2} (r - 3)^{2} + 6 r - 18 + 10 ln ∣ r - 3 ∣ + C

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What is the integral of (r^2+1)/(r-3) ?
The integral of (r^2+1)/(r-3) is 1/2 (r-3)^2+6r-18+10ln|r-3|+C