# I'm really having a problem integrating this equation, infact i have no idea

#### Anony111

1. The problem statement, all variables and given/known data

Please tell me how to integrate this...

2. Relevant equations

[(2x + 4)][(2x^2 + 3x + 1)^(1/2)]

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

#### HallsofIvy

$2x^2+ 3x+ 1= (2x+1)(x+1) Complete the square in the square root: [itex]2x^2+ 3x+ 1= 2(x^2+ (3/2)x+ 9/16)+1-9/8= 2(x+ 3/2)^2- 1/8$. Let u= x+ 3/2, then du= dx and x= u- 3/2 so that 2x+ 4= 2u+ 1 so the function to be integrated becomes $(2u+1)(2u^2+ 7/16)^{1/2}= 2\sqrt{2}u(u^2+ 7/32)^{1/2}+ \sqrt{2}(u^2+ 7/32}^{1/2}$. The first can be integrated by the substitution $v= u^2+ 7/32$ and the second by a trig substitution.

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#### Anony111

I get the idea but you have made one mistake; it should be 9/8 instead of 9/16 outside the bracket. Thanks alot though

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