Question

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

Answer 1

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 = e^z

⇒ dx = e^z 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