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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Expand

\frac{- 2 u + ( u + 1 ) ^{3} + 1}{u} : - 2 + \frac{( u + 1 ) ^{3}}{u} + \frac{1}{u}

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

\int 2 d u = 2 u

\int \frac{( u + 1 ) ^{3}}{u} d u = \frac{u ^{3}}{3} + \frac{3 u ^{2}}{2} + 3 u + ln ∣ u ∣

\int \frac{1}{u} d u = ln ∣ u ∣

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

Substitute back

u = x - 1

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

Simplify

- 2 (x - 1) + \frac{( x - 1 ) ^{3}}{3} + \frac{3 ( x - 1 ) ^{2}}{2} + 3 (x - 1) + ln ∣ x - 1 ∣ + ln ∣ x - 1 ∣ : x - 1 + \frac{1}{3} (x - 1)^{3} + \frac{3}{2} (x - 1)^{2} + 2 ln ∣ x - 1 ∣

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

Add a constant to the solution

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

\frac{1}{( s + 4 ) ^{2}}

y^{^{''}} - 2 y^{^{'}} + y = t + 1, y (0) = 0, y^{^{'}} (0) = 0

\frac{d}{d x} (5^{x + 1})

- 3 x^{2}

\frac{d}{d x} (\frac{4}{4 x + 1})

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