What is the integral of (x^3-4x^2+3)/(x-3) ?

The integral of (x^3-4x^2+3)/(x-3) is ((x-3)^3)/3+(5(x-3)^2)/2+3(x-3)-6ln|x-3|+C

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integral of (x^3-4x^2+3)/(x-3)

\int \frac{x ^{3} - 4 x ^{2} + 3}{x - 3} d x

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

Solution steps

\int \frac{x ^{3} - 4 x ^{2} + 3}{x - 3} d x

Apply u-substitution

= \int \frac{( u + 3 ) ^{3} - 4 ( u + 3 ) ^{2} + 3}{u} d u

Long division

\frac{( u + 3 ) ^{3} - 4 ( u + 3 ) ^{2} + 3}{u} : u^{2} + 5 u + 3 - \frac{6}{u}

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

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

\int 5 u d u = \frac{5 u ^{2}}{2}

\int 3 d u = 3 u

\int \frac{6}{u} d u = 6 ln ∣ u ∣

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

Substitute back

u = x - 3

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

Add a constant to the solution

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

What is the integral of (x^3-4x^2+3)/(x-3) ?
The integral of (x^3-4x^2+3)/(x-3) is ((x-3)^3)/3+(5(x-3)^2)/2+3(x-3)-6ln|x-3|+C