Integration by substitution
How would I Integrate this expression ?
http://imgur.com/6Ve5GQd
I tried to make u= 2x +1 But I get stuck with
(ln u)/ (squareroot u) at the end
http://imgur.com/6Ve5GQd
I tried to make u= 2x +1 But I get stuck with
(ln u)/ (squareroot u) at the end
Original post by jon2016
How would I Integrate this expression ?
http://imgur.com/6Ve5GQd
I tried to make u= 2x +1 But I get stuck with
(ln u)/ (squareroot u) at the end
http://imgur.com/6Ve5GQd
I tried to make u= 2x +1 But I get stuck with
(ln u)/ (squareroot u) at the end
Have you tried IBP with u = ln x and v = 1/sqrt(x)?
Original post by B_9710
The substitution works really nicely here.
Would you mind showing your working out ? Think im having a brain freeze
Original post by Zacken
Have you tried IBP with u = ln x and v = 1/sqrt(x)?
ah didnt occur to me to combine the two methods
I have attempted it but it got sort of messy, could I see your working out ?
Original post by jon2016
Would you mind showing your working out ? Think im having a brain freeze
It comes out to which should tell you something about the derivative of .
Original post by B_9710
It comes out to which should tell you something about the derivative of .
when you differentiate
Unparseable latex formula:
u = \ \sqrt{2x+1}}
you get
du = (2x+1)^-1/2 dx right ?
Then du/ (2x+1)^-1/2 = dxso you can now replace dx in the original expression with the above.So I have Ln(2x+1)/u * du/ (2x+1)^-1/2
(edited 7 years ago)
Original post by jon2016
when you differentiate
you get
du = (2x+1)^-1/2 dx right ?
Unparseable latex formula:
u = \ \sqrt{2x+1}}
you get
du = (2x+1)^-1/2 dx right ?
Yes.
Original post by B_9710
It comes out to which should tell you something about the derivative of .
what would be the final answer ?
Original post by jon2016
what would be the final answer ?
Use IBP with u = ln x and dv = 1
Quick Reply
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