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Sum

Evaluate : `int_2^3 (1)/(x^2 + 5x + 6)*dx`

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#### Solution

`int_2^3 (1)/(x^2 + 5x + 6)*dx`

= `int_2^3 (1)/((x + 2)(x + 3))*dx`

= `int_2^3 ((x + 3) - (x + 2))/((x + 2)(x + 3))*dx`

= `int_2^3 [1/(x + 2) - 1/(x + 3)]*dx`

= `[log (x + 2) - log(x + 3)]_2^3`

= `[log |(x + 2)/(x + 3)|]_2^3`

= `log((3 + 2)/(3 + 3)) - log((2 + 2)/(2 + 3))`

= `log (5)/(6) - log (4)/(5)`

= `log(5/6 xx 5/4)`

= `log(25/24)`.

Concept: Fundamental Theorem of Integral Calculus

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