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Question
Evaluate the integral by using substitution.
`int_0^1 x/(x^2 +1)`dx
Solution
Let `int_0^1 x/(x^2 + 1) dx`
Put x2 + 1 = t
2x dx = dt
When x =1, t = 2; x = 0, t = 1
`therefore I = int_1^2 dt/t`
∴ `I = 1/2 int_1^2 dt/t = [1/2 log t]_1^2`
`= 1/2 [log 2 - log 1]`
`= 1/2 log 2`
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