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Question
`int "dx"/(("x" - 8)("x" + 7))`=
Options
`1/15 log |("x" + 2)/("x" - 1)| + "c"`
`1/15 log |("x" + 8)/("x" + 7)| + "c"`
`1/15 log |("x"- 8)/("x" + 7)| + "c"`
(x − 8)(x − 7) + c
`1/15 log |("x" + 2)/("x"+ 1)| + "c"`
(x − 8)(x + 7) + c
Solution
`bb(1/15 log |("x"- 8)/("x" + 7)| + "c")`
Explanation:
I = `int "A"/("x" - 8) + "B"/("x" + 7)"dx"`
1 = A(x + 7) + B(x − 8)
When x = 8, A = `1/15` and x = −7, B = `(-1)/15`
∴ I = `int 1/15 (1/("x" - 8)) "dx" + int (-1)/15 (1/("x "+ 7)) "dx"`
= `int 1/15 log ("x" - 8)"dx" - int 1/15 log ("x" + 7)`
= `int 1/15 {log (("x" - 8)/("x" + 7))} + "c"`
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