# 2 integrals - - (within next 30 mins if possible)

1. Feb 26, 2008

### Mattofix

2 integrals - URGENT - (within next 30 mins if possible)

1. The problem statement, all variables and given/known data

$$\int_0^{b}\frac{dx}{x+\sqrt x}$$

$$\int$$$$^{b}_{}$$ (x + $$\sqrt{x})^{-1}$$dx
$$_{ 1}$$

3. The attempt at a solution

For the first i have tried x=t^2 but ended up with something i cant integrate (2t/(t^2 + t) dt

for the second i tried u= 1 + x^3 but that did'nt help -

Last edited: Feb 26, 2008
2. Feb 26, 2008

### morphism

Hint:
$$\frac{2t}{t^2 + t} = \frac{2t+1}{t^2 + t} - \frac{1}{t^2 + t}$$

This screams substitution for the first term and partial fractions for the second.

Your second integrand is exactly like your first one. Am I missing something?

3. Feb 26, 2008

### Mattofix

ahhhh- yeah the second is

$$\int$$$$^{\infty}_{}$$ (1 + x$$^{3}$$)$$^{-1/2}$$ dx
$$_{ 1}$$

4. Feb 26, 2008

### olgranpappy

try the substitution

y=sqrt[x]

...oh, right... that is what you tried... then factor out a y from numerator and denominator...

5. Feb 26, 2008

### Mattofix

cool - cheers

6. Feb 26, 2008

### motx

$$\int_0^{b}\frac{dx}{x+\sqrt x}$$

This thing works out nicely if you factor out $$\sqrt{x}$$, giving you

$$\sqrt{x}\left(\sqrt{x} + 1\right)$$

Then let u = $$\left(\sqrt{x} + 1\right)$$ and you're almost done.

7. Feb 26, 2008

### sutupidmath

8. Feb 26, 2008

### Mattofix

yeah - i got the x= t^2 but i could't integrate (2t/(t^2 + t) dt

thanks though

9. Feb 26, 2008

### sutupidmath

i also simplified it in there, now you will end up with 2 ln(1+t) on the interval sqrt b, and 0