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Question
Evaluate the definite integral:
`int_0^2 (6x +3)/(x^2 + 4)` dx
Solution
Let `I = int_0^2 (6x + 3)/(x^2 + 4) dx`
`= int_0^2 (6x) /(x^2 + 4) dx + int_0^2 3/ (x^2 + 4) dx`
`= 3 int_0^2 (2x)/(x^2 + 4) dx + [3 xx 1/2 tan^-1 x/2]_0^2`
Let `I_1 = 3 int_0^2 (2x)/ (x^2 + 4) dx`
Put x2 + 4 = t
⇒ 2x dx = dt
When x = 0
⇒ t = 4
When x = 2
⇒ t = 8
`= 3 int_4^8 dt/t = [3 log t]_4^8`
= 3 (log 8 - log 4)
= 3 log 2
⇒ `I = 3 log 2 + 3/2 [tan^-1 1 tan^-1 0]`
`= 3 log 2 + 3/2 xx pi/4 = 3 log 2 + (3pi)/8`
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