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C3 Integration with Substitution problem

Question:

Using u2=x+1
Solve integral 1/x(x+1)1/2 from 3 to 8..

The answer is ln3/2

I've tried many ways to solve this, but haven't managed to obtain the correct answer.

Thanks in advance guys. :smile:
Reply 1
Avatar for ezioaudi77
ezioaudi77
Original post by Aklima(:
Question:

Using u2=x+1
Solve integral 1/x(x+1)1/2 from 3 to 8..

The answer is ln3/2

I've tried many ways to solve this, but haven't managed to obtain the correct answer.

Thanks in advance guys. :smile:


What did you get as du/dx?
Reply 2
Avatar for LeeMrLee
LeeMrLee
u^2 = x+1

rearrange for x and you get x = u^2 - 1

sub both of these in the integral, and you get:

I = 1/[(u^2 - 1)u] = 1/(u+1)(u-1)u

split into partial fractions and then you'll get the logs.

2du/dx=1 => dx = 2du


EDIT: forgot the du/dx
(edited 11 years ago)
Reply 3
Avatar for MathMan
MathMan
6
Do you know implicit differentiation?
Reply 4
Avatar for Aklima(:
Aklima(:
OP
Original post by LeeMrLee
u^2 = x+1

rearrange for x and you get x = u^2 - 1

sub both of these in the integral, and you get:

I = 1/[(u^2 - 1)u] = 1/(u+1)(u-1)u

split into partial fractions and then you'll get the logs.

2du/dx=1 => dx = 2du


EDIT: forgot the du/dx


Ohh you use partial fractions for this question? Maybe that's why I couldn't do it.. Haven't been taught how to do that yet.. :colondollar: Thanks anyways! :smile:
Reply 5
Avatar for Aklima(:
Aklima(:
OP
Original post by MathMan
Do you know implicit differentiation?


No unfortunately, I think i picked out a C4 question which I thought was C3. :biggrin:
What paper is this question from? And which board


Posted from TSR Mobile
Reply 7
Avatar for MathMan
MathMan
6
Original post by Aklima(:
No unfortunately, I think i picked out a C4 question which I thought was C3. :biggrin:


Yeah I think you did, this question is screaming out implicit differentiation
Reply 8
Avatar for matematicas
matematicas
∫ 1/x((x+1)^ 1/2) dx

let u^2 =x+1

d(u^2)/dx=d(x+1)/dx

2u(du/dx) =1

2u du = dx

u^2 = x+1 then x = u^2 -1

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

2/ (u+1)(u-1) = A/(u+1) + B/(u-1)

A= -1 B= 1

∫ 2/ (u^2 -1) du = ∫ 1/(u-1) -1/(u+1) du = ln(u-1) - ln(u+1)

substitute back in for (x+1)^1/2 = u

ln(u-1) - ln(u+1) = ln(((x+1)^1/2) -1) - ln (((x+1)^1/2) +1)

put values of x in

(ln 2- ln 4) - ( ln 1 - ln 3) = ln (3/2)

(edited 11 years ago)
Reply 9
Avatar for TenOfThem
TenOfThem
18
Original post by MathMan
Yeah I think you did, this question is screaming out implicit differentiation


no need for it
Reply 10
Avatar for LeeMrLee
LeeMrLee
Original post by Aklima(:
Ohh you use partial fractions for this question? Maybe that's why I couldn't do it.. Haven't been taught how to do that yet.. :colondollar: Thanks anyways! :smile:



yeah partial fractions are the easiest way to do it (in terms of difficulty)

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