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

The integral of 1/(sqrt(1+\sqrt{1+x))} is 4(1/3 (1+sqrt(1+x))^{3/2}-sqrt(1+\sqrt{1+x)})+C

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

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

4 (\frac{1}{3} (1 + 1 + x)^{\frac{3}{2}} - 1 + 1 + x) + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int 4 (v^{2} - 1) d v

= \int 4 (v^{2} - 1) d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int v^{2} - 1 d v

Apply the Sum Rule:

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

= 4 (\int v^{2} d v - \int 1 d v)

\int v^{2} d v = \frac{v ^{3}}{3}

\int 1 d v = v

= 4 (\frac{v ^{3}}{3} - v)

Substitute back

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

Simplify

4 \frac{( 1 + 1 + x ) ^{3}}{3} - 1 + 1 + x : 4 (\frac{1}{3} (1 + 1 + x)^{\frac{3}{2}} - 1 + 1 + x)

= 4 (\frac{1}{3} (1 + 1 + x)^{\frac{3}{2}} - 1 + 1 + x)

Add a constant to the solution

= 4 (\frac{1}{3} (1 + 1 + x)^{\frac{3}{2}} - 1 + 1 + x) + C

3 x^{2} - 6

\int (x - 1)^{2} e^{x} d x

\int tan (2 θ) d θ

(y^{4} - 2 x y) d x + 3 x^{2} d y = 0

\frac{d y}{d x} = \frac{8 e ^{4 x}}{2 cos ( 2 y )}

What is the integral of 1/(sqrt(1+\sqrt{1+x))} ?
The integral of 1/(sqrt(1+\sqrt{1+x))} is 4(1/3 (1+sqrt(1+x))^{3/2}-sqrt(1+\sqrt{1+x)})+C