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

The integral of (x^2+2x-1)/(3-x) is -1/2 (3-x)^2+24-8x-14ln|3-x|+C

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

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

- \frac{1}{2} (3 - x)^{2} + 24 - 8 x - 14 ln ∣ 3 - x ∣ + C

Solution steps

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

Apply u-substitution

= \int - \frac{u ^{2} - 8 u + 14}{u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{u ^{2} - 8 u + 14}{u} d u

Expand

\frac{u ^{2} - 8 u + 14}{u} : u - 8 + \frac{14}{u}

= - \int u - 8 + \frac{14}{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{14}{u} d u)

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

\int 8 d u = 8 u

\int \frac{14}{u} d u = 14 ln ∣ u ∣

= - (\frac{u ^{2}}{2} - 8 u + 14 ln ∣ u ∣)

Substitute back

u = 3 - x

= - (\frac{( 3 - x ) ^{2}}{2} - 8 (3 - x) + 14 ln ∣ 3 - x ∣)

Simplify

- (\frac{( 3 - x ) ^{2}}{2} - 8 (3 - x) + 14 ln ∣ 3 - x ∣) : - \frac{1}{2} (3 - x)^{2} + 24 - 8 x - 14 ln ∣ 3 - x ∣

= - \frac{1}{2} (3 - x)^{2} + 24 - 8 x - 14 ln ∣ 3 - x ∣

Add a constant to the solution

= - \frac{1}{2} (3 - x)^{2} + 24 - 8 x - 14 ln ∣ 3 - x ∣ + C

What is the integral of (x^2+2x-1)/(3-x) ?
The integral of (x^2+2x-1)/(3-x) is -1/2 (3-x)^2+24-8x-14ln|3-x|+C