Question

integrate integration of (3x+1)√4x^2+12x+5 dx​

Answer 1

To Integrate :

∫(3x+1)√4x^2+12x+5 dx​

Solution :

∫(3x+1)√4x²+12x+5 dx​ is of the form P√Q

•So we'll express P in the form of Q such that

3x+1 = A( d(4x²+12x+5 )/dx ) + B

3x+1 = A( 8x + 12 ) + B ___(1)

3x+1 = 8Ax + 12A + B

3x+1 = 8Ax + (12A + B )

•By comparing 8A = 3 & 12A + B = 1

A = 3/8 & 12(3/8)+B = 1

A = 3/8 & 9/2 + B = 1

A = 3/8 & B = -7/2

•Hence ,

3x+1 = 3/8( 8x + 12 ) -7/2 ___(from 1)

=> ∫(3x+1)√4x²+12x+5 dx​ =

[ 3/8( 8x + 12 ) -7/2 ]√4x²+12x+5 dx​

= 3/8∫(8x+12)√4x²+12x+5 dx​

-7/2 ∫√4x²+12x+5 dx​ _____(2)

•Now , 3/8∫(8x+12)√4x²+12x+5 dx​ =

let 4x²+12x+5 = t

(8x+12)dx = dt

=> 3/8∫(8x+12)√4x²+12x+5 dx​ =

3/8 ∫√t dt

= 3/8( t^³/² )/(3/2)+ K

= ( t^³/² )/4 + K

=[ (4x²+12x+5)³/² ] / 4 + K ____(3)

•And , -7/2 ∫√4x²+12x+5 dx​ =

Taking 4 common

= -7/2×2∫√(x²+3x+5/4) dx​

Adding and subtracting 4/4

= -7 ∫√[ (x²+3x+5/4+4/4) -4/4 ] dx

= -7 ∫√[ (x²+3x+9/4) - 1 ]

= -7 ∫√[ (x + 3/2)² - 1² ]

= -7[ (x + 3/2)/2 √[ (x + 3/2)² - 1² ] -1/2ln |(x + 3/2) + √[ (x + 3/2)² - 1² ] | + M _____(4)

•Putting (3) & (4) in (1)

=> ∫(3x+1)√4x²+12x+5 dx​ =

[ (4x²+12x+5)³/² ] / 4 -7[ (x + 3/2)/2 √[ (x + 3/2)² - 1² ] -1/2ln |(x + 3/2) + √[ (x + 3/2)² - 1² ] | + C

• Hence ,∫(3x+1)√4x²+12x+5 dx​ =

[ (4x²+12x+5)³/² ] / 4 -7[ (x + 3/2)/2 √[ (x + 3/2)² - 1² ] -1/2ln |(x + 3/2) + √[ (x + 3/2)² - 1² ] | + C