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Question
Evaluate the following : `int_0^1 1/(1 + sqrt(x))*dx`
Sum
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Solution
Let I = `int_0^1 1/(1 + sqrt(x))*dx`
Put `sqrt(x)` = t
∴ x = t2 and dx = 2t·dt
When x = 0, t = 0
When x = , t = 1
∴ I = `int_0^1 1/(1 + t)2t*dt`
= `2 int_0^1 t/(1 + t)*dt`
= `2 int_0^1 ((1 + t) - 1)/(1 + t)*dt`
= `2 int_0^1 (1 - 1/(1 + t))*dt`
= `2[t - log|1 + t|]_0^1`
= `2[1 - log2 - 0 + log1]`
= 2(1 - log 2) ...[∵ log 1 = 0]
= 2 – 2log 2
= 2 – log 4.
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