# integrating help

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#1
Hi im having a problem with this question:
(iii) Show that the integral of 1/(4x^2 -1)^0.5 (with lower bound 0.625 and upper bound 1.3) = 0.5(ln(5)-ln(2))

I took out a multiple of 1/2, leaving me to integrate 1/(x^2 -0.25)^0.5 to get 0.5ln(x+(x^2 -0.25)) . However, I keep getting an answer of 0.5(ln(5/2) -ln(1)) when I sub in 1.3 and 0.625. The mark scheme hasn't taken the 1/2 out and has instead integrated to get 0.5ln(2x+(4x^2 -1)^0.5) which I understand. I just don't understand why my method isn't working.
Any help would be appreciated
0
3 weeks ago
#2
(Original post by lilybeth647321)
Hi im having a problem with this question:
(iii) Show that the integral of 1/(4x^2 -1)^0.5 (with lower bound 0.625 and upper bound 1.3) = 0.5(ln(5)-ln(2))

I took out a multiple of 1/2, leaving me to integrate 1/(x^2 -0.25)^0.5 to get 0.5ln(x+(x^2 -0.25)) . However, I keep getting an answer of 0.5(ln(5/2) -ln(1)) when I sub in 1.3 and 0.625. The mark scheme hasn't taken the 1/2 out and has instead integrated to get 0.5ln(2x+(4x^2 -1)^0.5) which I understand. I just don't understand why my method isn't working.
Any help would be appreciated

Use the "laws of logs" to simplify your answer. Don't forget ln(1) is 0.
Last edited by ghostwalker; 3 weeks ago
2
#3
(Original post by ghostwalker)

Use the "laws of logs" to simplify your answer. Don't forget ln(1) is 0.
of course! Thanks for helping out 0
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