What is the integral of 1/(sqrt(9+\sqrt{4+x))} ?

The integral of 1/(sqrt(9+\sqrt{4+x))} is 4(1/3 (9+sqrt(4+x))^{3/2}-9sqrt(9+\sqrt{4+x)})+C

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integral of 1/(sqrt(9+\sqrt{4+x))}

\int \frac{1}{9 + 4 + x} d x

4 (\frac{1}{3} (9 + 4 + x)^{\frac{3}{2}} - 9 9 + 4 + x) + C

Solution steps

\int \frac{1}{9 + 4 + x} d x

Apply u-substitution

: \int \frac{1}{9 + u} d u

= \int \frac{1}{9 + u} d u

Apply u-substitution

: \int \frac{2 v}{9 + v} d v

= \int \frac{2 v}{9 + v} d v

Take the constant out:

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

= 2 \cdot \int \frac{v}{9 + v} d v

Apply u-substitution

: \int 2 (w^{2} - 9) d w

= 2 \cdot \int 2 (w^{2} - 9) d w

Take the constant out:

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

= 2 \cdot 2 \cdot \int w^{2} - 9 d w

Apply the Sum Rule:

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

= 2 \cdot 2 (\int w^{2} d w - \int 9 d w)

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

\int 9 d w = 9 w

= 2 \cdot 2 (\frac{w ^{3}}{3} - 9 w)

Substitute back

= 2 \cdot 2 \frac{( 9 + 4 + x ) ^{3}}{3} - 9 9 + 4 + x

Simplify

2 \cdot 2 \frac{( 9 + 4 + x ) ^{3}}{3} - 9 9 + 4 + x : 4 (\frac{1}{3} (9 + 4 + x)^{\frac{3}{2}} - 9 9 + 4 + x)

= 4 (\frac{1}{3} (9 + 4 + x)^{\frac{3}{2}} - 9 9 + 4 + x)

Add a constant to the solution

= 4 (\frac{1}{3} (9 + 4 + x)^{\frac{3}{2}} - 9 9 + 4 + x) + C

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What is the integral of 1/(sqrt(9+\sqrt{4+x))} ?
The integral of 1/(sqrt(9+\sqrt{4+x))} is 4(1/3 (9+sqrt(4+x))^{3/2}-9sqrt(9+\sqrt{4+x)})+C