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

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

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

\int \frac{x}{x + 1 - ( x + 1 )} d x

- x - 1 - 2 x + 1 + C

Solution steps

\int \frac{x}{x + 1 - ( x + 1 )} d x

Factor

x + 1 - (x + 1) : - (- x + 1 + (x + 1))

= \int \frac{x}{- ( - x + 1 + x + 1 )} d x

Take the constant out:

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

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

Apply u-substitution

: \int 2 (u + 1) d u

= - \int 2 (u + 1) d u

Take the constant out:

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

= - 2 \cdot \int u + 1 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

= - 2 (\int u d u + \int 1 d u)

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

\int 1 d u = u

= - 2 (\frac{u ^{2}}{2} + u)

Substitute back

u = x + 1

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

Simplify

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

= - x - 1 - 2 x + 1

Add a constant to the solution

= - x - 1 - 2 x + 1 + C

\frac{\partial}{\partial y} (x^{2} e^{y z^{2}})

\int_{0}^{1} x 1 - x d x

\frac{3}{2 3 x + 13}, at x = - 3

x, \frac{d ^{2} x}{d t ^{2}} + x = (t^{2} + t) sin (t)

(- 5, 11), (- 10, 6)

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