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

The integral of 1/(sqrt(\sqrt{x+1))} is 4/3 (x+1)^{3/4}+C

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

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

\frac{4}{3} (x + 1)^{\frac{3}{4}} + C

Solution steps

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

Apply u-substitution

Apply the Power Rule

: \frac{4}{3} u^{\frac{3}{4}}

= \frac{4}{3} u^{\frac{3}{4}}

Substitute back

u = x + 1

= \frac{4}{3} (x + 1)^{\frac{3}{4}}

Add a constant to the solution

= \frac{4}{3} (x + 1)^{\frac{3}{4}} + C

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What is the integral of 1/(sqrt(\sqrt{x+1))} ?
The integral of 1/(sqrt(\sqrt{x+1))} is 4/3 (x+1)^{3/4}+C