Integration by substitution where square root is U^2

Daveami
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Hi there,

I am having difficulty with one aspect of intergration by substitution where the substituion of a square root is U^2, wondering if anyone can help.

Problem:

Integral of: 2x√(3x-4) dx by substituting U^2 = 3x-4

Would du^2/dx = 3 therefore 1/3 du^2 = dx (I think this is where I am going wrong)


Im coming out with an answer of: 2/45(9x+8)(3x-4)^3/2 + k

However the answer in the book is: 4/135(9x+8)(3x-4)^3/2 + k

Any help would be greatly appreciated.

Regards

Dave
 
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Daveami said:
Integral of: 2x√(3x-4) dx by substituting U^2 = 3x-4

Would du^2/dx = 3

Your error is that the derivative of u^2 isn't du^2.

u^{2}=3x-4

2u du=3dx

du=\frac{3}{2\sqrt{3x-4}}dx
 
Last edited:
Ah brilliant! Thanks for the help mate!
 
And really, the substitution is u = sqrt(3x - 4). If u >= 0, this is equivalent to u^2 = 3x - 4.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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