If ∫ 1 2 d x x 2 - 2 x + 4 3 2 = k k + 5 , then k is equal to

If ∫12dxx2-2x+432=kk+5, then k is equal to
  1. 4
  2. 2
  3. 3
  4. 1

Solution

∫ 1 2 dx ( ( x−1 ) 2 +3 ) 3 2

Substituting x-1=3tan⁡θ​

⇒dx=3sec2θ·dθ​

Now ∫0π63 sec 2 θ​ ⁡dθ​33 sec 3 θ​

=13∫0π6cosθdθ​=13sin⁡θ0π6=16

⇒16=kk+5=k+5=6k

=k=1.

Asked in: JEE Main 2017 (09 Apr Online)

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