What is the integral from-2 to 1 of (x^2+4x+8)/(x-2) ?

The integral from-2 to 1 of (x^2+4x+8)/(x-2) is -40ln(2)+33/2

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integral from-2 to 1 of (x^2+4x+8)/(x-2)

\int_{- 2}^{1} \frac{x ^{2} + 4 x + 8}{x - 2} d x

- 40 ln (2) + \frac{33}{2}

Decimal Notation

- 11.22588 \dots

Solution steps

\int_{- 2}^{1} \frac{x ^{2} + 4 x + 8}{x - 2} d x

Apply u-substitution

: \int_{- 4}^{- 1} \frac{u ^{2} + 8 u + 20}{u} d u

= \int_{- 4}^{- 1} \frac{u ^{2} + 8 u + 20}{u} d u

Expand

\frac{u ^{2} + 8 u + 20}{u} : u + 8 + \frac{20}{u}

= \int_{- 4}^{- 1} u + 8 + \frac{20}{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_{- 4}^{- 1} u d u + \int_{- 4}^{- 1} 8 d u + \int_{- 4}^{- 1} \frac{20}{u} d u

\int_{- 4}^{- 1} u d u = - \frac{15}{2}

\int_{- 4}^{- 1} 8 d u = 24

\int_{- 4}^{- 1} \frac{20}{u} d u = - 40 ln (2)

= - \frac{15}{2} + 24 - 40 ln (2)

Combine the fractions

24 - \frac{15}{2} : \frac{33}{2}

= - 40 ln (2) + \frac{33}{2}

g (x) = cos (x) + 1, f (x) = sin (x)

\int_{1}^{9} 3 - x d x

f (x) = x^{3} (x - 5)^{2}

x \to 9 - lim (\frac{∣ x - 9 ∣}{x - 9})

What is the integral from-2 to 1 of (x^2+4x+8)/(x-2) ?
The integral from-2 to 1 of (x^2+4x+8)/(x-2) is -40ln(2)+33/2