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99 Percentile Qs Bank for JEE MainMathematicsIndefinite Integration

If ∫ 1 ( x - 2 ) 5 ( x - 1 ) 4 d x = ∑ r = - 4 - 1 A r x - 2 x - 1 r + ∑ r = 1 3 A r x - 2 x - 1 r + B f x , then f x =

Options

  1. Alog ( x - 2 ) - log ( x - 1 )
  2. Bx - 2 x - 1 + log x
  3. Cx + log x - 2 x - 1
  4. Dlog x

Correct answer

A. log ( x - 2 ) - log ( x - 1 )

Step-by-step solution

I = ∫ 1 x - 2 5 . x - 1 4 d x ⇒   I = ∫ d x x - 2 9 . x - 1 x - 2 4 Let x - 1 x - 2 = t ⇒   x - 2 . 1 - x - 1 . 1 x - 2 2 d x = d t ⇒   - 1 x - 2 2 d x = d t       . . . 1 Also, x - 1 x - 2 = t ⇒   x - 1 = t x - 2 t ⇒   x = 2 t - 1 t - 1 ⇒   x - 2 = 1 t - 1     . . . 2 ∴   I = ∫ - d t 1 t - 1 7 t 4 ⇒   I = ∫ 1 - t 7 t 4 d t ⇒   I = ∫ 1 - c 1 7 t + c 2 7 t 2 - c 3

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