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Question
Solve the following : `int_3^5 dx/(sqrt(x + 4) + sqrt(x - 2)`
Solution
Let I = `int_3^5 dx/(sqrt(x + 4) + sqrt(x - 2)`
= `int_3^5 (1)/(sqrt(x + 4) + sqrt(x - 2)) xx (sqrt(x + 4) - sqrt(x - 2))/(sqrt(x + 4) - sqrt(x - 2))*dx`
= `int_3^5 (sqrt(x + 4) - sqrt(x - 2))/((sqrt(x + 4))^2 - (sqrt(x - 2))^2)*dx`
= `int_3^5 (sqrt(x + 4) - sqrt(x - 2))/(x + 4 - (x - 2))*dx`
= `int_3^5 (sqrt(x + 4) - sqrt(x - 2))/(6)*dx`
= `(1)/(6) int_3^5 (x + 4)^(1/2)*dx - (1)/(6) int_3^5 (x - 2)^(1/2)*dx`
= `(1)/(6) [((x + 4)^(3/2))/(3/2)]_3^5 - (1)/(6)[((x - 2)^(3/2))/(3/2)]_3^5`
= `(1)/(9)[(9)^(3/2) - (7)^(3/2)] - (1)/(9) [(3)^(3/2) - (1)^(3/2)]`
= `(1)/(9) (27 - 7sqrt(7)) - (1)/(9) (3sqrt(3) - 1)`
= `(1)/(9)(27 - 7sqrt(7) - 3sqrt(3) + 1)`
∴ I = `(1)/(9)(28 - 3sqrt(3) - 7sqrt(7))`.
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