[samd]

Integrate:

1/[1 + (x - 1)^0.5] with respect to x, using a substitution of u^2 = x - 1.

According to the back of the book: between the limits of 5 and 2 (for x) you should get 2 + 2ln(2/3) = 1.19 (2dp).

It is in Exercise 6F, on page 99 of Heinemann Edexcel C4, Question 3 part (c).

Any help would be much appreciated.

Thank you.

[m:)ckel]

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

x=2, u=1
x=5, u=2

=> INT (2u/1+u)du , from 1 to 2

= 2 INT [(1+u)-1 / (1+u)] du

= 2 INT [1 - 1/(1+u)] du

= 2[u - ln(1+u)] , from 1 to 2

= 2[(2-ln3) - (1-ln2)]

= 2[1 + ln(2/3)]

= 2 + 2ln(2/3)

[J.F.N]

u^2=x-1 --> dx/du=2u
Plugging in, you want to integrate
2u/(1+u) with respect to u
split this into u/(1+u) + u/(1+u)
These integrate into
(u+1) - ln(u+1) + (u+1) - ln(u+1)=2(u+1)-2ln(u+1)
since we want to integrate from x=2 to x=5, we want u from 1 to 2
plugging in, we get 2(3)-2ln(3)-2(2)+2ln(2)= 2 + 2(ln(2/3))=1.19

[samd]

Originally Posted by mockel

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

x=2, u=1
x=5, u=2

=> INT (2u/1+u)du , from 1 to 2

= 2 INT [(1+u)-1 / (1+u)] du

= 2 INT [1 - 1/(1+u)] du

= 2[u - ln(1+u)] , from 1 to 2

= 2[(2-ln3) - (1-ln2)]

= 2[1 + ln(2/3)]

= 2 + 2ln(2/3)

Thank you to both of you.

I have highlighted the step that I couldn't do. More practice needed I guess.