Question

integration of DX/(2x^2+X+3)​

Answer 1

Step-by-step explanation:

To find --->

∫ dx / (2x² + x + 3 )

Solution---> First of all we make whole square in denominator as follows

2x² + x + 3 = 2 ( x² + x / 2 + 3 / 2 )

= 2 { x² + 2 ( 1 / 4 ) x + 3 / 2 }

Adding and subtracting (1 / 4 )² to form whole square

= 2 [{ x² + 2(1/4)x +(1/4)²} +3/2 - (1/4 )²]

We have an identity

( a + b )² = a² + b² + 2 ab , we get

= 2 { ( x + 1/4 )² + (3 / 2 ) - ( 1 / 16 ) }

= 2 { ( x + 1 / 4 )² + ( 24 - 1 ) / 16 }

= 2 { ( x + 1 / 4 )² + 23 / 16 }

= 2 { ( x + 1 / 4 )² + ( √23 / 4 )² }

Now returning to original problem

∫ dx / (2x² + x + 3 )

Putting value of denominator from above we get

=∫ dx / ( x + 1/4 )² + ( √23 / 4 )²

Let x + 1/4 = t

dx = dt

=∫ dt /{ t² + (√23 / 4 )² }

= 1/ ( √23 /4) tan⁻¹ { t / (√23 / 4 ) } + C

Putting value of t we get

= 4 / √23 tan⁻¹ { 4 ( x + 1/4 ) / √23 } + C

= 4 / √23 tan⁻¹ { ( 4x + 1 ) / √23 } + C