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

The integral of (sqrt(x-2))/(x+1) is 2sqrt(3)(-arctan(sqrt(1/3 (x-2)))+sqrt(1/3 (x-2)))+C

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integral of (sqrt(x-2))/(x+1)

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

23 (- arctan (\frac{1}{3} (x - 2)) + \frac{1}{3} (x - 2)) + C

Solution steps

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

Apply u-substitution

: \int \frac{2 u ^{2}}{u ^{2} + 3} d u

= \int \frac{2 u ^{2}}{u ^{2} + 3} d u

Take the constant out:

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

= 2 \cdot \int \frac{u ^{2}}{u ^{2} + 3} d u

\frac{u ^{2}}{u ^{2} + 3} = - \frac{3}{u ^{2} + 3} + 1

= 2 \cdot \int - \frac{3}{u ^{2} + 3} + 1 d u

Apply Integral Substitution

= 2 \cdot \int 3 (- \frac{1}{v ^{2} + 1} + 1) d v

Take the constant out:

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

= 23 \cdot \int - \frac{1}{v ^{2} + 1} + 1 d v

Apply the Sum Rule:

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

= 23 (- \int \frac{1}{v ^{2} + 1} d v + \int 1 d v)

\int \frac{1}{v ^{2} + 1} d v = arctan (v)

\int 1 d v = v

= 23 (- arctan (v) + v)

Substitute back

= 23 (- arctan (\frac{x - 2}{3}) + \frac{x - 2}{3})

Simplify

23 (- arctan (\frac{x - 2}{3}) + \frac{x - 2}{3}) : 23 (- arctan (\frac{1}{3} (x - 2)) + \frac{1}{3} (x - 2))

= 23 (- arctan (\frac{1}{3} (x - 2)) + \frac{1}{3} (x - 2))

Add a constant to the solution

= 23 (- arctan (\frac{1}{3} (x - 2)) + \frac{1}{3} (x - 2)) + C

\frac{\partial}{\partial x} ((x - 3)^{3})

\frac{d}{d x} (x^{4} - 5 + x^{- 2} + 41 x^{- 4})

y^{^{''}} + 4 y = 2 sin (2 t)

\frac{d}{d x} (11 + 3 s + 5 y, 2 + 5 + 2 y, 3 + 2 x + y)

x \to 0 lim ({x})

What is the integral of (sqrt(x-2))/(x+1) ?
The integral of (sqrt(x-2))/(x+1) is 2sqrt(3)(-arctan(sqrt(1/3 (x-2)))+sqrt(1/3 (x-2)))+C