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Evaluate: aexbexaexbex∫aex+be-xaex-be−x dx

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प्रश्न

Evaluate: `int ("ae"^("x") + "be"^(-"x"))/("ae"^("x") - "be"^(−"x"))` dx

मूल्यांकन
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उत्तर

Let I = `int ("ae"^("x") + "be"^(-"x"))/("ae"^("x") - "be"^(−"x"))` dx

Put aex − be−x = t 

∴ `["ae"^("x") − "be"^(−"x") .(-1)] "dx" = "dt"`

∴ `("ae"^("x") + "be"^(−"x")) "dx" = "dt"`

∴ I = `int "dt"/"t"`

∴ I = `int 1/"t" "dt"`

∴ I = log | t | + c

∴ I = log | aex − be−x | + c

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पाठ 5: Integration - MISCELLANEOUS EXERCISE - 5 [पृष्ठ १३८]

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बालभारती Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
पाठ 5 Integration
MISCELLANEOUS EXERCISE - 5 | Q IV. 2) ii) | पृष्ठ १३८

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