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

The integral of ((x+sqrt(2x-5))/(x-2)) is x+2ln(2|x-2|)+2sqrt(2x-5)-2arctan(sqrt(2x-5))-2+C

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integral of ((x+sqrt(2x-5))/(x-2))

\int (\frac{x + 2 x - 5}{x - 2}) d x

x + 2 ln (2 ∣ x - 2 ∣) + 2 2 x - 5 - 2 arctan (2 x - 5) - 2 + C

Solution steps

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

Apply u-substitution

: \int \frac{u + 5 + 2 u}{2 ( u + 1 )} d u

= \int \frac{u + 5 + 2 u}{2 ( u + 1 )} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{u + 5 + 2 u}{u + 1} d u

Expand

\frac{u + 5 + 2 u}{u + 1} : \frac{u}{u + 1} + \frac{5}{u + 1} + \frac{2 u}{u + 1}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int \frac{u}{u + 1} d u + \int \frac{5}{u + 1} d u + \int \frac{2 u}{u + 1} d u)

\int \frac{u}{u + 1} d u = u + 1 - ln ∣ u + 1 ∣

\int \frac{5}{u + 1} d u = 5 ln ∣ u + 1 ∣

\int \frac{2 u}{u + 1} d u = 4 (- arctan (u) + u)

= \frac{1}{2} (u + 1 - ln ∣ u + 1 ∣ + 5 ln ∣ u + 1 ∣ + 4 (- arctan (u) + u))

Substitute back

u = 2 x - 5

= \frac{1}{2} (2 x - 5 + 1 - ln ∣ 2 x - 5 + 1 ∣ + 5 ln ∣ 2 x - 5 + 1 ∣ + 4 (- arctan (2 x - 5) + 2 x - 5))

Simplify

\frac{1}{2} (2 x - 5 + 1 - ln ∣ 2 x - 5 + 1 ∣ + 5 ln ∣ 2 x - 5 + 1 ∣ + 4 (- arctan (2 x - 5) + 2 x - 5)) : x + 2 ln (2 ∣ x - 2 ∣) + 2 2 x - 5 - 2 arctan (2 x - 5) - 2

= x + 2 ln (2 ∣ x - 2 ∣) + 2 2 x - 5 - 2 arctan (2 x - 5) - 2

Add a constant to the solution

= x + 2 ln (2 ∣ x - 2 ∣) + 2 2 x - 5 - 2 arctan (2 x - 5) - 2 + C

\int \frac{4 x + 5}{( x ^{2} - 2 x + 2 ) ^{\frac{3}{2}}} d x

\int (t - 2)^{2} d t

\int tan (c)

\int \frac{x}{x ^{4} + 9} d x

\int \frac{3}{x ^{\frac{1}{3}}} d x

What is the integral of ((x+sqrt(2x-5))/(x-2)) ?
The integral of ((x+sqrt(2x-5))/(x-2)) is x+2ln(2|x-2|)+2sqrt(2x-5)-2arctan(sqrt(2x-5))-2+C