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

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

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

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

- y - 1 - 2 y + 1 + C

Solution steps

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

Factor

y + 1 - (y + 1) : - (- y + 1 + (y + 1))

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

Take the constant out:

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

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

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 = y + 1

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

Simplify

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

= - y - 1 - 2 y + 1

Add a constant to the solution

= - y - 1 - 2 y + 1 + C

\int \frac{1}{e ^{x} + e ^{- x}} d x

\frac{1}{n + k e ^{n}}

f (x) = 3 (x)^{3} + cos (x)

\int sin^{2020} (d) x d x

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