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प्रश्न
If `int_2^e [1/logx - 1/(logx)^2].dx = a + b/log2`, then ______.
विकल्प
a = e, b = –2
a = e, b = 2
a = –e, b = 2
a = –e, b = –2
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उत्तर
If `int_2^e [1/logx - 1/(logx)^2].dx = a + b/log2`, then a = e, b = –2.
Explanation:
Given that, `int_2^e [1/logx - 1/(logx)^2].dx = a + b/log2`
Put logx = z
⇒ x = ez
⇒ dx = ez dz
∴ `int_2^e [1/logx - 1/(logx)^2].dx = int_log2^1 [1/z - 1/z^2]e^z.dz`
= `int_log2^1 e^z [1/z + d(1/z)].dz`
= `[e^z . 1/z]_log2^1`
= `e - 2/log2`
∴ a = e and b = –2
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