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

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

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

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

- ln y + 1 + 1 + ln y + 1 - 1 + 2 y + 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{2 u ^{2}}{u ^{2} - 1} d u

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

Factor

u^{2} - 1 : - (- u^{2} + 1)

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

Take the constant out:

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

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

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

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

\int \frac{1}{- u ^{2} + 1} d u = \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2}

\int 1 d u = u

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

Substitute back

u = y + 1

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

Simplify

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

= - ln y + 1 + 1 + ln y + 1 - 1 + 2 y + 1

Add a constant to the solution

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

\frac{d}{d x} (e^{x} cos (x^{2}))

y \to - 1 lim (- y^{5} - 3 y^{4})

f (x) = e^{2 x}, (0, 1)

n = 1 \sum \infty (- 1)^{n} n^{- 1}

y^{^{''}} + 5 y^{^{'}} = e^{- t}, y (0) = 0, y^{^{'}} (0) = 0

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