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

The integral of (sqrt(x+1)-1)/(sqrt(x+1)+1) is x+4ln|sqrt(x+1)+1|-4sqrt(x+1)-4+C

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

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

x + 4 ln x + 1 + 1 - 4 x + 1 - 4 + C

Solution steps

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

Expand

\frac{x + 1 - 1}{x + 1 + 1} : \frac{x + 1}{x + 1 + 1} - \frac{1}{x + 1 + 1}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int \frac{x + 1}{x + 1 + 1} d x = x + 2 ln x + 1 + 1 - 2 x + 1 - 2

\int \frac{1}{x + 1 + 1} d x = - 2 ln x + 1 + 1 + 2 x + 1 + 2

= x + 2 ln x + 1 + 1 - 2 x + 1 - 2 - (- 2 ln x + 1 + 1 + 2 x + 1 + 2)

Simplify

x + 2 ln x + 1 + 1 - 2 x + 1 - 2 - (- 2 ln x + 1 + 1 + 2 x + 1 + 2) : x + 4 ln x + 1 + 1 - 4 x + 1 - 4

= x + 4 ln x + 1 + 1 - 4 x + 1 - 4

Add a constant to the solution

= x + 4 ln x + 1 + 1 - 4 x + 1 - 4 + C

y = 1 9^{x}

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

x \to 0 - lim (\frac{2}{3} x^{- \frac{1}{3}})

\int_{0}^{\infty} e^{- 14 x} d x

\frac{d ^{3}}{d x ^{3}} (\frac{1 - 5 x}{1 + 5 x})

What is the integral of (sqrt(x+1)-1)/(sqrt(x+1)+1) ?
The integral of (sqrt(x+1)-1)/(sqrt(x+1)+1) is x+4ln|sqrt(x+1)+1|-4sqrt(x+1)-4+C