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

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

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

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

x + 2 ln x + 1 + 1 - 2 x + 1 - 2 + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Expand

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

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

\int 2 d u = 2 u

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

Substitute back

u = 1 + 1 + x

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

Simplify

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

= x + 2 ln x + 1 + 1 - 2 x + 1 - 2

Add a constant to the solution

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

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