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

The integral of 1/(sqrt(x)\sqrt{\sqrt{x)+1}} is 4sqrt(\sqrt{x)+1}+C

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

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

4 x + 1 + C

Solution steps

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

Apply u-substitution

: \int \frac{2}{u + 1} d u

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

Apply u-substitution

: \int 2 d v

= 2 \cdot \int 2 d v

Integral of a constant:

\int a d x = a x

= 2 \cdot 2 v

Substitute back

= 2 \cdot 2 x + 1

Simplify

= 4 x + 1

Add a constant to the solution

= 4 x + 1 + C

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What is the integral of 1/(sqrt(x)\sqrt{\sqrt{x)+1}} ?
The integral of 1/(sqrt(x)\sqrt{\sqrt{x)+1}} is 4sqrt(\sqrt{x)+1}+C