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

[Mattofix]

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

Homework Statement

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

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

The Attempt at a Solution

For the first i have tried x=t^2 but ended up with something i can't integrate (2t/(t^2 + t) dt

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

 

[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?

 

[Mattofix]

ahhhh- yeah the second is

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

 

[olgranpappy]

try the substitution

y=sqrt[x]

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

 

[Mattofix]

cool - cheers

 

[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.

 

[sutupidmath]

Homework Statement

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

-

I did almost the whole thing on another post of yours!

https://www.physicsforums.com/showthread.php?t=217839

 

[Mattofix]

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

thanks though

 

[sutupidmath]

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

thanks though

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