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Question
`int_0^1 1/(2x + 5) dx` = ______.
Options
`1/2` log `7/5`
`1/2` log `5/7`
log `7/5`
`1/4` log `7/5`
Solution
`int_0^1 1/(2x + 5) dx` = `bb(underline(1/2 log 7/5))`.
Explanation:
⇒ 2x + 5 = t
⇒ 2dx = dt
⇒ dx = `1/2`dt
⇒ `int_5^7 1/2 (dt)/t`
⇒ `1/2 (logt)_5^7`
⇒ `1/2 [log7 - log5]`
⇒ `(log7 - log5)/2`
⇒ `1/2 [log 7 - log 5]`
⇒ `1/2 log 7/5` .....`[log m - log n = log m/n]`
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