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

The integral of (x^2+2x-5)/(x-3) is ((x-3)^2)/2+8(x-3)+10ln|x-3|+C

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

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

\frac{( x - 3 ) ^{2}}{2} + 8 (x - 3) + 10 ln ∣ x - 3 ∣ + C

Solution steps

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

Apply u-substitution

= \int \frac{u ^{2} + 8 u + 10}{u} d u

Expand

\frac{u ^{2} + 8 u + 10}{u} : u + 8 + \frac{10}{u}

= \int u + 8 + \frac{10}{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 8 d u + \int \frac{10}{u} d u

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

\int 8 d u = 8 u

\int \frac{10}{u} d u = 10 ln ∣ u ∣

= \frac{u ^{2}}{2} + 8 u + 10 ln ∣ u ∣

Substitute back

u = x - 3

= \frac{( x - 3 ) ^{2}}{2} + 8 (x - 3) + 10 ln ∣ x - 3 ∣

Add a constant to the solution

= \frac{( x - 3 ) ^{2}}{2} + 8 (x - 3) + 10 ln ∣ x - 3 ∣ + C

What is the integral of (x^2+2x-5)/(x-3) ?
The integral of (x^2+2x-5)/(x-3) is ((x-3)^2)/2+8(x-3)+10ln|x-3|+C