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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Expand

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

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

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

\int 2 d u = 2 u

\int \frac{1}{u} d u = ln ∣ u ∣

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

Substitute back

u = 1 + 1 + x

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

Simplify

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

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

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\int \frac{x}{3 x ^{2} + 6} d x

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