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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{( u + 1 ) ^{2}}{u} d u

= \int \frac{( u + 1 ) ^{2}}{u} d u

Expand

\frac{( u + 1 ) ^{2}}{u} : u + 2 + \frac{1}{u}

= \int u + 2 + \frac{1}{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{1}{u} d u

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

\int 2 d u = 2 u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{u ^{2}}{2} + 2 u + ln ∣ u ∣

Substitute back

u = x - 1

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

Add a constant to the solution

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

\int (x^{2} - 6)^{4} (2 x) d x

x \to \infty lim (x (x - 1))

y = (5 x - 2)^{3}

\frac{x - 7}{x + 7}

x \to - 1 lim (3 - x^{2})

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