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BITSAT2014MathematicsIndefinite IntegrationActual

Evaluate: 1 1+3 ² x+8 ² x dx

Options

  1. A1 6 ⁻¹(2 x )+ C
  2. B⁻¹(2 x )+ C
  3. C1 6 ⁻¹ ( 2 x 3 )+ C
  4. DNone of these

Correct answer

C. 1 6 ⁻¹ ( 2 x 3 )+ C

Step-by-step solution

I = 1 1+3 ² x +8 ² x dx Dividing the numerator and denominator by ² x , we get I= ² x ² x+3 ² x+8 d x= ² x 4 ² x+9 d x Putting x=t ² x d x=d t, we get I = dt 4 t ²+9 = 1 4 dt t ²+(3 / 2)² = 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

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