What is the integral of 1/((1+sqrt(x))^{3/2)} ?

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

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integral of 1/((1+sqrt(x))^{3/2)}

\int \frac{1}{( 1 + x ) ^{\frac{3}{2}}} d x

2 (2 1 + x + \frac{2}{1 + x}) + C

Solution steps

\int \frac{1}{( 1 + x ) ^{\frac{3}{2}}} d x

Apply u-substitution

= \int \frac{2 u}{( 1 + u ) ^{\frac{3}{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{u}{( 1 + u ) ^{\frac{3}{2}}} d u

Apply u-substitution

= 2 \cdot \int \frac{v - 1}{v ^{\frac{3}{2}}} d v

Expand

\frac{v - 1}{v ^{\frac{3}{2}}} : \frac{1}{v ^{\frac{1}{2}}} - \frac{1}{v ^{\frac{3}{2}}}

= 2 \cdot \int \frac{1}{v ^{\frac{1}{2}}} - \frac{1}{v ^{\frac{3}{2}}} 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

= 2 (\int \frac{1}{v ^{\frac{1}{2}}} d v - \int \frac{1}{v ^{\frac{3}{2}}} d v)

\int \frac{1}{v ^{\frac{1}{2}}} d v = 2 v^{\frac{1}{2}}

\int \frac{1}{v ^{\frac{3}{2}}} d v = - \frac{2}{v ^{\frac{1}{2}}}

= 2 (2 v^{\frac{1}{2}} - (- \frac{2}{v ^{\frac{1}{2}}}))

Substitute back

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

Simplify

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

= 2 (2 1 + x + \frac{2}{1 + x})

Add a constant to the solution

= 2 (2 1 + x + \frac{2}{1 + x}) + C

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