Question

integrate 1/(sqrtx(4x+1))

Answer 1

try u^2 = 4x+1

Answer 2

Is the equation:
\[\frac{1}{\sqrt{x(4x+1)}}\]?

Answer 3

Oh looks i missed that x in front when I looked at it earlier :o

Answer 4

I think it is, I am not sure

Answer 5

yes that is the equation

Answer 6

oh wait no only the first x is under the square root

Answer 7

\[\int\frac{\mathrm dx}{\sqrt x(4x+1)}\]

Answer 8

yess unklerhaukus that is the correct equation

Answer 9

Why is it \(dx\) and not 1?

Answer 10

it's the same thing

Answer 11

Oh, thanks for the clarrification.

Answer 12

try the substitution \(u=\sqrt x\)
\(\mathrm d u = \,?\,\mathrm dx\)

Answer 13

I don't think that works...du=1/2x^(-1/2) ??

Answer 14

\[\mathrm du =\frac12x^{-1/2} \mathrm dx\]

that is correct , but

if you write it another way

\[\mathrm du = \frac{\mathrm dx}{2\sqrt x}\]

Answer 15

and you have \(\dfrac{\mathrm dx}{\sqrt x}\) in your integrand already, so it will cancel out nicely

Answer 16

What do you get for the integral in-terms of u and du

Answer 17

im not sure what to sub for the (4x+1)

Answer 18

we choose \(u = \sqrt x\)
so \(u^2 =x\)

Answer 19

du/(4u^2+1)

Answer 20

?

Answer 21

and then would i use partial fractions?

Answer 22

careful, dx/√x = 2du

Answer 23

oh yeah i took the 2 outside the integrand

Answer 24

i think you might find a similar integral in a table

Answer 25

\[\boxed{\displaystyle\int\frac{\mathrm dx}{a^2+x^2} = \frac1a\arctan\tfrac xa+c}\]

Answer 26

you'll need you manipulate the integrand a bit until you have this form

Answer 27

@fgmel88 , how are you going with it?