हिंदी

If ∫2e[1logx-1(logx)2].dx=a+blog2, then ______.

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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

MCQ
रिक्त स्थान भरें
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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

shaalaa.com
Methods of Evaluation and Properties of Definite Integral
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2025-2026 (March) Model set 2 by shaalaa.com

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