1. ## evaluate the integral

integrate 1/(square root x + x^3/2)

2. Originally Posted by twilightstr
integrate 1/(square root x + x^3/2)
Is it $\frac{1}{\sqrt{x} + x^{3/2}} ?$ If so, start by substituting $u = \sqrt{x} \Rightarrow dx = 2 \sqrt{x} \, du$.

3. no, entire the denominator is square root

4. Originally Posted by twilightstr
no, entire the denominator is square root
Then please make the effort to use appropriate grouping symbols.

$\int\frac{dx}{\sqrt{x + x^{3/2}}}$

$=\int\frac{dx}{\sqrt{x(1 + x^{1/2})}}$

$=\int\frac{dx}{\sqrt x\sqrt{1 + x^{1/2}}}$

$=\int\frac{x^{-1/2}\,dx}{\sqrt{1 + x^{1/2}}}$

$=2\int\frac{(1/2)x^{-1/2}\,dx}{\sqrt{1 + x^{1/2}}}$

You should be able to get it from here. Use $u=1+x^{1/2}.$