What is the integral of (2x^2+7x-3)/(x-2) ?

The integral of (2x^2+7x-3)/(x-2) is x^2+11x+19ln|x-2|-26+C

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integral of (2x^2+7x-3)/(x-2)

\int \frac{2 x ^{2} + 7 x - 3}{x - 2} d x

x^{2} + 11 x + 19 ln ∣ x - 2 ∣ - 26 + C

Solution steps

\int \frac{2 x ^{2} + 7 x - 3}{x - 2} d x

Apply u-substitution

: \int \frac{2 u ^{2} + 15 u + 19}{u} d u

= \int \frac{2 u ^{2} + 15 u + 19}{u} d u

Expand

\frac{2 u ^{2} + 15 u + 19}{u} : 2 u + 15 + \frac{19}{u}

= \int 2 u + 15 + \frac{19}{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 2 u d u + \int 15 d u + \int \frac{19}{u} d u

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

\int 15 d u = 15 u

\int \frac{19}{u} d u = 19 ln ∣ u ∣

= u^{2} + 15 u + 19 ln ∣ u ∣

Substitute back

u = x - 2

= (x - 2)^{2} + 15 (x - 2) + 19 ln ∣ x - 2 ∣

Simplify

(x - 2)^{2} + 15 (x - 2) + 19 ln ∣ x - 2 ∣ : x^{2} + 11 x + 19 ln ∣ x - 2 ∣ - 26

= x^{2} + 11 x + 19 ln ∣ x - 2 ∣ - 26

Add a constant to the solution

= x^{2} + 11 x + 19 ln ∣ x - 2 ∣ - 26 + C

ln (arctan (x))

(t + e^{t}) (6 - t)

y^{^{''}} + 5 y^{^{'}} + 6 y = 6

\frac{d}{d t} (t^{2})

y = e^{- s i n (x)}

What is the integral of (2x^2+7x-3)/(x-2) ?
The integral of (2x^2+7x-3)/(x-2) is x^2+11x+19ln|x-2|-26+C