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

The integral of 1/(sqrt(x)\sqrt{1+x)} is 2ln|sqrt(x+1)+sqrt(x)|+C

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integral of 1/(sqrt(x)\sqrt{1+x)}

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

2 ln x + 1 + x + C

Solution steps

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

Apply u-substitution

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

Use the common integral:

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

= 2 ln u^{2} + 1 + u

Substitute back

u = x

= 2 ln (x)^{2} + 1 + x

(x)^{2} = x

= 2 ln x + 1 + x

Add a constant to the solution

= 2 ln x + 1 + x + C

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