1. ## Integration by substitution

Hey, having a bit of trouble with a substitution problem:

Integrate : x/(2x+1)^1/2 dx using the substitution method

I get 1/3(2x+1)^3/2 - (2x+1)^1/2 + c

The answer is down as 1/3(x-1)((2x+1)^1/2) + c

I cant see how to get to this, any help would be greatly appreciated...

2. What is your working? It seems sensible to pick u=2x+1 and through algebra x = (u-1)/2. From the sub we know that du = 2 dx or dx = du/2

x/(2x+1)^(1/2) dx = (u-1)/(2u^(1/2)) * du/2 = 1/4 * [(u-1)/u^(1/2)] du

We can simplify that up a bit to give

$\dfrac{1}{4}\int\left(u^{1/2}-u^{-1/2}\right)du$

Those are relatively simple integrals

3. Integration by parts works nicely here...

Let u = x so that du = dx, and let dv = (2x + 1)^(-1/2)dx so that v = (2x + 1)^(1/2)

Then int[x(2x + 1)^(-1/2)dx] = x(2x + 1)^(1/2) - int[(2x + 1)^(1/2)dx].

I'm sure you can go from here.

4. Hello, jlaing!

Integrate: .x/(2x+1)^{1/2} dx

If the expression under the radical is linear, let $\,u$ equal the entire radical.

. . . . . . . . . _____
We have: .√2x + 1 .= .u . , 2x + 1 .= .u^2

. . . . . . x .= .(u^2 - 1)/2 . . dx .= .u du

. . . . . . . . . . . . . .(u^2 - 1)/2
Substitute: .[INT] -------------- (u du) . = . (1/2) [INT] {u^2 - 1) du
. . . . . . . . . . . . . . . . . u

Got it?

5. Thats great thank you, it seems I can do the actual integration part but I have something basic going wrong when I am trying to get the final answer.

For example:

= (1/4)( (2/5)u5/2 - (2/3)u3/2 ) I can get to this line but I cant get to the next

= ( (2x + 1)3/2(3x - 1) ) / 15 + K

I know it is something basic but I cant see it

6. you should get y= 1/6(2x+1)^1.5 - 0.5(2x+1)^0.5 + c

7. Originally Posted by jlaing
Thats great thank you, it seems I can do the actual integration part but I have something basic going wrong when I am trying to get the final answer.

For example:

= (1/4)( (2/5)u5/2 - (2/3)u3/2 ) I can get to this line but I cant get to the next

= ( (2x + 1)3/2(3x - 1) ) / 15 + K

I know it is something basic but I cant see it
Use ^ to denote "to the power of" otherwise what you write is too confusing to be easily read.