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KCET2023MathematicsIndefinite Integration

1 1+3 ^2 x+8 ^2 x d x is equals to

Options

  1. A⁻¹ ( 2 x 3 )+C
  2. B1 6 ⁻¹ ( 2 x 3 )+C
  3. C6 ⁻¹ ( 2 x 3 )+C
  4. D1 6 ⁻¹(2 x)+C

Correct answer

B. 1 6 ⁻¹ ( 2 x 3 )+C

Step-by-step solution

Let I= 1 1+3 ^2 x+8 ^2 x d x Dividing the numerator and denominator by ^2 x , we get aligned & I= ^2 x ^2 x+3 ^2 x+8 d x & I= ^2 x 1+ ^2 x+3 ^2 x+8 d x & I= ^2 x 4 ^2 x+9 d x aligned Putting, x=t ^2 x d x=d t , we get gathered I= d t 4 t^2+9 = 1 4 d t t^2+ ( 3 2 )^2 +C I= 1 4 1 3 / 2 ⁻¹ ( t 3 / 2 )+C I= 1 6 ⁻¹ ( 2 t 3 )+C= 1 6 ⁻¹ ( 2 x 3 )+C gathered

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