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

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

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

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

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

Solution steps

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

Apply u-substitution

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

Expand

\frac{u ^{2} + 2 u + 2}{u} : u + 2 + \frac{2}{u}

= \int u + 2 + \frac{2}{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 u d u + \int 2 d u + \int \frac{2}{u} d u

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

\int 2 d u = 2 u

\int \frac{2}{u} d u = 2 ln ∣ u ∣

= \frac{u ^{2}}{2} + 2 u + 2 ln ∣ u ∣

Substitute back

u = x - 1

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

Add a constant to the solution

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

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