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

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SUMMARY

The discussion focuses on the integration of the expression \((2x + 4)\sqrt{2x^2 + 3x + 1}\). The participant successfully factors the quadratic \(2x^2 + 3x + 1\) as \((2x + 1)(x + 1)\) and completes the square to transform the integrand. They utilize the substitution \(u = x + \frac{3}{2}\) and derive a new expression for integration, identifying the need for both substitution and trigonometric methods for solving the integral. A correction is noted regarding the constant term outside the bracket, which should be \(9/8\) instead of \(9/16\).

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Homework Statement



Please tell me how to integrate this...

Homework Equations




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

The Attempt at a Solution


 
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2x^2+ 3x+ 1= (2x+1)(x+1)<br /> Complete the square in the square root: 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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I get the idea but you have made one mistake; it should be 9/8 instead of 9/16 outside the bracket. Thanks a lot though
 

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