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# integration using substitution watch

1. integrate using substitution
x(2x+1)^(1/2) dx u=2x +1

= x- u^(1/2) du/2
= (u-1)/2 u- u^(1/2) du/2
= 1/4 [(u-1) - U^(1/2) du
=1/4 [ -u^(3/2) = u^(1/2)
=1/4 [-2/5u^(5/2) + 2/3u^(3/2)]
2. You have not shown us what you have tried
3. (Original post by vjosa)
integrate using substitution
x(2x+1)^(1/2) dx u=2x+1

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

3x^2 (x^3 - 2)^4 dx u=x^3 - 2

12xe^(3x^2) dx u=3x^2

final answers in terms of x
4. (Original post by vjosa)
integrate using substitution
x(2x+1)^(1/2) dx u=2x +1

= x* u^(1/2) du/2
= (u-1)/2 u* u^(1/2) du/2
= 1/4 [(u-1)*[ u^(1/2)] du
=1/4 [ -u^(3/2) = u^(1/2)
=1/4 [-2/5u^(5/2) + 2/3u^(3/2)]
You were doing fine till you got to:=1/4 [ -u^(3/2) = u^(1/2)

It should be 1/4*Integral of [ u^(3/2)- u^(1/2)] //You add the powers//
You were doing fine till you got to:=1/4 [ -u^(3/2) = u^(1/2)

It should be 1/4*Integral of [ u^(3/2)- u^(1/2)] //You add the powers//

thanks

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Updated: November 29, 2011
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