I was doing this problem and was having some problems understanding what I should, as there seems to be an x value still remaining.

∫√1 + (√35×7)2dx\int{\sqrt{1\ + \ (\frac{\sqrt{35x}}{7})^2}} dx

Used integration by substitution method:

I made u=√35x7u = \frac{\sqrt{35x}}{7} , which inturn resulted in du=3514√35xdxdu = \frac{35}{14\sqrt{35x}} dx

∫√1 + (u)214√35x35du\int{\sqrt{1\ + \ (u)^2}} \frac{14\sqrt{35x}}{35}du

And this is where I am stuck, I seem to still have an x value.

I know that simplifying the problem ahead of time, would resolve the complexity of the problem, but I wanted to see whether I could still get a solution if I did not simplify.

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Can you express √x\sqrt {x} in terms of uu ? (You never said what your uu is.)

– Ian

Oct 21 at 2:57

Try u=1+(√35x/7)2u=1+\left(\sqrt{35x}/7\right)^2. In general, try to add any “nearby” constants to your choice of uu.

– team-erdos

Oct 21 at 2:59

@Ian I have updated my answer to reflect what u is.

– user36278

Oct 21 at 3:05

Your choice of uu means that a multiple of √x\sqrt{x} is a (different) multiple of uu. So you have C∫√1+u2uduC \int \sqrt{1+u^2} u du for some (explicit) constant CC. In general when doing say ∫f1(x)f2(x)dx\int f_1(x) f_2(x) dx by substitution, you may be able to write f1(x)dxf_1(x) dx as f3(u)duf_3(u) du, but then you still need to be able to represent f2(x)f_2(x) as some f4(u)f_4(u), which may require some additional algebra (based on your original definition of uu).

– Ian

2 days ago

For example, ∫x√x+1dx=∫x√udu\int x \sqrt{x+1} dx = \int x \sqrt{u} du where u=x+1u=x+1, but then x=u−1x=u-1 so you really have ∫(u−1)√udu\int (u-1) \sqrt{u} du. That intermediate step where you wrote an integral with both xx and uu should be regarded as scratch work, not a proper step (but it is useful scratch work).

– Ian

2 days ago

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