Calculus by parts - urgent help for exam please

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Tayeeba

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Calculus integration by substitution - urgent help for exam please

Hello, my professor did this solution on board, but i am having a confusion. The part I marked with arrow - is it right? I think it would be x/2 instead of 2x in the denominator

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So, if I understand you correctly, you're confused as to why/how the denominator of the integral goes from \(\displaystyle x \cdot \sqrt{4x+1}\) to \(\displaystyle 2x \cdot \sqrt{x+\dfrac{1}{4}}\)

Assuming that's correct, let's start by factoring out a 4 to get \(\displaystyle x \cdot \sqrt{4 \left(x + \dfrac{1}{4} \right)}\). We know that, for any real numbers (or expressions comprised thereof), \(\displaystyle \sqrt{ab}=\sqrt{a} \cdot \sqrt{b}\). Hence, we can rewrite the expression as: \(\displaystyle x \cdot \sqrt{4} \cdot \sqrt{x + \dfrac{1}{4}}\). We also know that 4 = 22, so that gives us the final form of \(\displaystyle 2x \cdot \sqrt{x+\dfrac{1}{4}}\)
 
ksdhart2 said:
So, if I understand you correctly, you're confused as to why/how the denominator of the integral goes from \(\displaystyle x \cdot \sqrt{4x+1}\) to \(\displaystyle 2x \cdot \sqrt{x+\dfrac{1}{4}}\)

Assuming that's correct, let's start by factoring out a 4 to get \(\displaystyle x \cdot \sqrt{4 \left(x + \dfrac{1}{4} \right)}\). We know that, for any real numbers (or expressions comprised thereof), \(\displaystyle \sqrt{ab}=\sqrt{a} \cdot \sqrt{b}\). Hence, we can rewrite the expression as: \(\displaystyle x \cdot \sqrt{4} \cdot \sqrt{x + \dfrac{1}{4}}\). We also know that 4 = 22, so that gives us the final form of \(\displaystyle 2x \cdot \sqrt{x+\dfrac{1}{4}}\)
Click to expand...

Thank you so very much for your explanation and for the time you took to explain it this much clearly, the God bless you.
 
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