Algebric Simplifying Process

I was just viewing a tutorial on how to solve the integral of ∫x × 6÷√x dx\int{x\ \times\ 6 \div\sqrt{x}}\ dx

And I noticed they simplified it to : ∫6√x dx\int6\sqrt{x}\ dx , before they started on the integral.

I wanted to know how they did such a simplification process.

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Write the root as an exponent of one half and use properties of exponents.
– O. Von Seckendorff
Oct 21 at 1:01

  

 

MathJax tip: You don’t have to \space. All you need is \ with an actual space following and it turns into a space automatically.
– Simple Art
Oct 21 at 1:09

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1 Answer
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x×6÷√x=6x√x=6√x⋅√x√x=6√xx \times 6 \div \sqrt x = \frac{6x}{\sqrt x} = \frac{6\sqrt x \cdot \sqrt x}{\sqrt x} = 6\sqrt x

  

 

Thanks for the clarification!
– user36278
Oct 21 at 1:03

  

 

Would it be possible to do the integral without simplifying it into 6(sqrt(x))?
– user36278
Oct 21 at 1:10

  

 

@user36278 Sure, you could’ve used integration by parts for example. Which is why we use algebraic simplification first
– Simple Art
Oct 21 at 1:10

  

 

@ Simple Art Wow, I just tried to do it with integration by parts and the answer looks really complicated, and I think I might have done it incorrectly as the graphs don’t match.
– user36278
Oct 21 at 1:26

  

 

@user36278 ∫6x⋅1√xdxu        dv+∫2√x⋅6dxv            du=6x⋅2√xu        v\underset{\color{red}{u \ \ \ \ \ \ \ \ dv}} {\int 6x \cdot \frac{1}{\sqrt x}dx } + \underset{\color{red}{v \ \ \ \ \ \ \ \ \ \ \ \ du}} {\int 2\sqrt x \cdot6 dx } = \underset{\color{red}{u \ \ \ \ \ \ \ \ v}} {6x\cdot 2\sqrt x }
– Deepak
Oct 21 at 2:06