# Easy Integral Question (1 Viewer)

#### ISAM77

##### Member
Hi guys, I'm stuck on something trivial.

letting x = (u-4)^2

I get to the correct solution in terms of u: 2u - 8ln|u| + c

However, my confusion begins when changing the answer back to be in terms of x. Since x = (u-4)^2 , u = x^(1/2)+4 OR u =-x^(1/2)+4

So shouldn't there be two solutions in terms of x? One for the positive case and one for the negative case? Solutions across the internet and in back of book give the positive answer only. Could someone please elaborate and clear this up for me.

#### liamkk112

##### Well-Known Member
Hi guys, I'm stuck on something trivial.

View attachment 42843 letting x = (u-4)^2

I get to the correct solution in terms of u: 2u - 8ln|u| + c

However, my confusion begins when changing the answer back to be in terms of x. Since x = (u-4)^2 , u = x^(1/2)+4 OR u =-x^(1/2)+4

So shouldn't there be two solutions in terms of x? One for the positive case and one for the negative case? Solutions across the internet and in back of book give the positive answer only. Could someone please elaborate and clear this up for me.
maybe think of it as letting sqrt(x) = u -4,
so then 1/2sqrt(x) dx= du
so dx = 2sqrt(x)du = 2(u-4)du
from here u get the same result but now the substitution back in makes sense as u = 4+sqrt(x)
so this is why the positive square root is the typical answer

alternatively, if u = 4-sqrt(x), then sqrt(x) = 4-u, if u sub this in then after some manipulation u get that the integral is -2u -8ln(8-u) +C
which becomes -2(4-sqrt(x))-8ln(4+sqrt(x)) +C = 2sqrt(x) - 8ln(4+sqrt(x)) +C after absorbing stuff into the constant
which is completly equivalent to when u do the u-sub as u = 4+sqrt(x)
so actually it does not matter, so long as u pick which sign at the beginning, though obviously the sub u = 4+sqrt(x) gives an easier result

the issue comes when u cross out sqrt((u-4)^2), this actually becomes |u-4|, in general you pick the positive case by "cancelling" it out without thinking about it but you then need to consider the positive case of the substitution only. this is why it's a better idea to just sub sqrt(x) = u-4, it removes the confusion with the absolute value

#### Drongoski

##### Well-Known Member
Hello Hello Hello.

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#### Drongoski

##### Well-Known Member
Hello again, clarifying my approach to integration..

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##### New Member
$\bg_white \int \frac{1}{4 + \sqrt{x} } \, dx$

Let $\bg_white u = x^{1/2} \Rightarrow u^2 = x$

\bg_white \begin{align*} u^2 &= x \\ \text{Differentiate both sides with respect to x} \\ \frac{d}{dx} (u^2) &= \frac{d}{dx}x \,\, \text{Apply chain rule to LHS} \\ \frac{d}{du} (u^2) \frac{du}{dx } &= 1 \\ 2u \frac{du}{dx} &= 1 \\ dx &= 2u\,du \\ \int \frac{2u}{4 + u} \,\, du \\ 2\int \frac{u + 4 - 4}{u + 4} \,\, du \\ 2\int 1 - \frac{4}{u + 4} \,\, du \\ 2\left[ u - 4\ln|u + 4| \right]+ C \\ 2\sqrt{x} - 8\ln|\sqrt{x} + 4| + C \end{align*}

#### ISAM77

##### Member
alternatively, if u = 4-sqrt(x), then sqrt(x) = 4-u, if u sub this in then after some manipulation u get that the integral is -2u -8ln(8-u) +C
which becomes -2(4-sqrt(x))-8ln(4+sqrt(x)) +C = 2sqrt(x) - 8ln(4+sqrt(x)) +C after absorbing stuff into the constant
which is completly equivalent to when u do the u-sub as u = 4+sqrt(x)
This reasoning mirrors my own thoughts, so thankyou for the clarity. This helped!

#### ISAM77

##### Member
Hello again, clarifying my approach to integration..
dx = d(sqrt(x))^2 , fine.

Why does d(sqrt(x))^2 = 2*sqrt(x)*d*sqrt(x) ?

As gorgeous as your apples may be. The integral of 1/x is not my source of confusion with your solution.

Why does dx^2 = 2xdx ?
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Edit, I got it:
dx^2 = d/dx * x^2 * dx = 2x dx

d(sqrt(x))^2 = d/d*sqrt(x) * sqrt(x)^2 * d*sqrt(x) = 2 * sqrt(x) * d*sqrt(x)

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Where did you learn this method?

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#### Drongoski

##### Well-Known Member
dx = d(sqrt(x))^2 , fine.

Why does d(sqrt(x))^2 = 2*sqrt(x)*d*sqrt(x) ?

As gorgeous as your apples may be. The integral of 1/x is not my source of confusion with your solution.

Why does dx^2 = 2xdx ?
---

Edit, I got it:
dx^2 = d/dx * x^2 * dx = 2x dx

d(sqrt(x))^2 = d/d*sqrt(x) * sqrt(x)^2 * d*sqrt(x) = 2 * sqrt(x) * d*sqrt(x)

---
Where did you learn this method?
I "invented" this method (let's for now call it the "Drongoski Method").

A 2-hour lesson would be sufficient for an introduction to my method. So far, I've attracted no interest. I think they don't know what my approach is all about. But I believe it to be a very powerful, intuitive and very efficient approach to Integration, and a useful supplement to how mechanical integration has been taught over the last 100 years in school and in textbooks. Would be a pity if not passed on.

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