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

The integral of ((3x^2+2))/(x-1) is (3(x-1)^2)/2+6(x-1)+5ln|x-1|+C

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

\int \frac{( 3 x ^{2} + 2 )}{x - 1} d x

\frac{3 ( x - 1 ) ^{2}}{2} + 6 (x - 1) + 5 ln ∣ x - 1 ∣ + C

Solution steps

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

Apply u-substitution

= \int \frac{3 u ^{2} + 6 u + 5}{u} d u

Expand

\frac{3 u ^{2} + 6 u + 5}{u} : 3 u + 6 + \frac{5}{u}

= \int 3 u + 6 + \frac{5}{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 u d u + \int 6 d u + \int \frac{5}{u} d u

\int 3 u d u = \frac{3 u ^{2}}{2}

\int 6 d u = 6 u

\int \frac{5}{u} d u = 5 ln ∣ u ∣

= \frac{3 u ^{2}}{2} + 6 u + 5 ln ∣ u ∣

Substitute back

u = x - 1

= \frac{3 ( x - 1 ) ^{2}}{2} + 6 (x - 1) + 5 ln ∣ x - 1 ∣

Add a constant to the solution

= \frac{3 ( x - 1 ) ^{2}}{2} + 6 (x - 1) + 5 ln ∣ x - 1 ∣ + C

What is the integral of ((3x^2+2))/(x-1) ?
The integral of ((3x^2+2))/(x-1) is (3(x-1)^2)/2+6(x-1)+5ln|x-1|+C