What is the integral of ((3x+10))/((x-2)) ?

The integral of ((3x+10))/((x-2)) is 3(x-2)+16ln|x-2|+C

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integral of ((3x+10))/((x-2))

integral

\frac{( 3 x + 10 )}{( x - 2 )}

3 (x - 2) + 16 ln ∣ x - 2 ∣ + C

Solution steps

\int \frac{3 x + 10}{x - 2} d x

Apply u-substitution

= \int \frac{3 u + 16}{u} d u

Expand

\frac{3 u + 16}{u} : 3 + \frac{16}{u}

= \int 3 + \frac{16}{u} d u

Apply the Sum Rule:

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

= \int 3 d u + \int \frac{16}{u} d u

\int 3 d u = 3 u

\int \frac{16}{u} d u = 16 ln ∣ u ∣

= 3 u + 16 ln ∣ u ∣

Substitute back

u = x - 2

= 3 (x - 2) + 16 ln ∣ x - 2 ∣

Add a constant to the solution

= 3 (x - 2) + 16 ln ∣ x - 2 ∣ + C

What is the integral of ((3x+10))/((x-2)) ?
The integral of ((3x+10))/((x-2)) is 3(x-2)+16ln|x-2|+C