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Question
`int_2^3 "x"/("x"^2 - 1)` dx = ____________.
Options
`(1/3) log (8/3)`
`(-1/3) log (8/3)`
`(1/2) log (8/3)`
`(-1/2) log (8/3)`
MCQ
Fill in the Blanks
Solution
`int_2^3 "x"/("x"^2 - 1)` dx = `underline((1/2) log (8/3))`.
Explanation:
We have, l = `int_2^3 "x"/("x"^2 - 1)` dx
Put, x2 − 1 = t = 2x dx = dt = xdx = `"dt"/2`
When, x = 2, t = 3 and x = 3, t = 8
∴ l = `1/2 int_3^8 "dt"/"t"`
l = `1/2 [log "t"]_3^8`
= `1/2 [log 8 - log 3]`
l = `1/2 log (8/3)`
shaalaa.com
Fundamental Theorem of Integral Calculus
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