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

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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{- u + 3}{u} d u

= \int \frac{- u + 3}{u} d u

Expand

\frac{- u + 3}{u} : - 1 + \frac{3}{u}

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

Apply u-substitution

: \int - \frac{3 v - 1}{v ^{2}} d v

= \int - \frac{3 v - 1}{v ^{2}} d v

Take the constant out:

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

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

Apply Integration By Parts

: - \frac{v - 1}{v} - \int - \frac{1}{2 v v - 1} d v

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

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

= - 3 (- \frac{v - 1}{v} - (- arctan (v - 1)))

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (sin (x y^{2}))

f (x) = - x^{2} + 8 x - 8

- 3 e^{t}

\frac{x ^{2} - 2}{e ^{2 x}}

x \to - 1 lim (lo g_{10} (x))

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