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

The integral of (x^3+3x-4)/(x-2) is 3x^2+3x+10ln|x-2|+1/3 (x-2)^3-18+C

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

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

3 x^{2} + 3 x + 10 ln ∣ x - 2 ∣ + \frac{1}{3} (x - 2)^{3} - 18 + C

Solution steps

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

Apply u-substitution

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

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

Expand

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

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

\int 3 d u = 3 u

\int \frac{( u + 2 ) ^{3}}{u} d u = \frac{u ^{3}}{3} + 3 u^{2} + 12 u + 8 ln ∣ u ∣

\int \frac{2}{u} d u = 2 ln ∣ u ∣

= 3 u + \frac{u ^{3}}{3} + 3 u^{2} + 12 u + 8 ln ∣ u ∣ + 2 ln ∣ u ∣

Substitute back

u = x - 2

= 3 (x - 2) + \frac{( x - 2 ) ^{3}}{3} + 3 (x - 2)^{2} + 12 (x - 2) + 8 ln ∣ x - 2 ∣ + 2 ln ∣ x - 2 ∣

Simplify

3 (x - 2) + \frac{( x - 2 ) ^{3}}{3} + 3 (x - 2)^{2} + 12 (x - 2) + 8 ln ∣ x - 2 ∣ + 2 ln ∣ x - 2 ∣ : 3 x^{2} + 3 x + 10 ln ∣ x - 2 ∣ + \frac{1}{3} (x - 2)^{3} - 18

= 3 x^{2} + 3 x + 10 ln ∣ x - 2 ∣ + \frac{1}{3} (x - 2)^{3} - 18

Add a constant to the solution

= 3 x^{2} + 3 x + 10 ln ∣ x - 2 ∣ + \frac{1}{3} (x - 2)^{3} - 18 + C

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\int y e^{2 x y} d x

\int \frac{1}{x \cdot 1 - ln ^{2} ( x )} d x

What is the integral of (x^3+3x-4)/(x-2) ?
The integral of (x^3+3x-4)/(x-2) is 3x^2+3x+10ln|x-2|+1/3 (x-2)^3-18+C