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

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

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

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

\frac{x}{x + 1} (x + 1) - ln x + 1 + x + C

Solution steps

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

Apply u-substitution

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

Apply Integration By Parts

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

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

= \frac{u - 1}{u} u - ln u + u - 1

Substitute back

u = x + 1

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

Simplify

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

Add a constant to the solution

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

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