What is the integral of (x^3-3x^2+5x-3)/(x-1) ?

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

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integral of (x^3-3x^2+5x-3)/(x-1)

\int \frac{x ^{3} - 3 x ^{2} + 5 x - 3}{x - 1} d x

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

Solution steps

\int \frac{x ^{3} - 3 x ^{2} + 5 x - 3}{x - 1} d x

Expand

\frac{x ^{3} - 3 x ^{2} + 5 x - 3}{x - 1} : \frac{x ^{3}}{x - 1} - \frac{3 x ^{2}}{x - 1} + \frac{5 x}{x - 1} - \frac{3}{x - 1}

Apply the Sum Rule:

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

= \int \frac{x ^{3}}{x - 1} d x - \int \frac{3 x ^{2}}{x - 1} d x + \int \frac{5 x}{x - 1} d x - \int \frac{3}{x - 1} d x

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

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

\int \frac{5 x}{x - 1} d x = 5 (x - 1 + ln ∣ x - 1 ∣)

\int \frac{3}{x - 1} d x = 3 ln ∣ x - 1 ∣

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

Simplify

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

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

Add a constant to the solution

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

y = 3 x - x^{2}, y = - 5 x

f (x) = 3 x + 1, at x = 5

\frac{d}{d x} (2 e^{6 x})

\int x 1 + x^{4} d x

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