How do you evaluate the integral?

#int 1/(sqrt(x+x^(3/2)))dx#

1 Answer
Oct 4, 2017

The answer is #=4sqrt(1+sqrtx)+C#

Explanation:

Perform this integral by substitutions

First substitution

The denominator is

#sqrt(1+x^(3/2))=sqrtxsqrt(1+sqrtx)#

Let #u=sqrtx#, #=>#, #du=dx/(2sqrtx)#

Therefore,

#int(dx)/(sqrtxsqrt(1+sqrtx))=int(2du)/(sqrt(1+u))#

Second substitution

Let #v=u+1#, #=>#, #dv=du#

Therefore,

#int(2du)/(sqrt(1+u))=int(2dv)/(sqrtv)#

#=(2sqrtv)/(1/2)#

#=4sqrtv#

#=4sqrt(1+u)#

#=4sqrt(1+sqrtx)+C#