Question

Need help with u-substitution

Answer 1

\int \left(1/sqrt\left(5\:+\:sqrt\left(x\right)\right)\right)
\[\int\limits_{}^{}\frac{ 1 }{ \sqrt{5+\sqrt{x}} }dx\]

Answer 2

Ok, I would first consider rewriting it, like (5+sqrt(x))^-1/2

Answer 3

Then what would be an appropriate u sub so we can make it easy?

Answer 4

in u-substitution we usually substitute the algebraic operation that hinders us most the calculation of the integral, so we want to remove it by substitution with hopes that the integral becomes simpler.
\[\int\limits_{}^{}\frac{ 1 }{\sqrt{5+ \sqrt x} }dx\]
We will take the substitution "u" as the square root on the denominator so we can get rid of it:
\[u= 5+\sqrt x \iff \frac{ du }{ dx }=\frac{ 1 }{ 2 \sqrt x } \iff dx=2 \sqrt x du\]
Now with that out of the way, let's solve for "x" on the same change of variable:
\[u= 5+ \sqrt x \iff \sqrt x = u-5 \iff x= (u-5)^2\]
replacing all the information:
\[\int\limits_{}^{}\frac{ 1 }{ \sqrt{5+ \sqrt x} }dx \iff \int\limits_{}^{}\frac{ 1 }{ \sqrt u }.2\sqrt x du\]
\[\int\limits_{}^{}\frac{ 2 \sqrt{(u-5)^2} }{ \sqrt u } du\]
\[\int\limits_{}^{}\frac{ 2(u-5) }{ \sqrt u }du\]
And now it's just a matter of solving this last expression, which I'll leave to you.

Answer 5

If you get rid of the sqrt in the denominator it may be easier... You'll have (5+sqrt(x))^-1/2 U sub 5+sqrt(x)
It becomes the integral of u ^-1/2

Answer 6

I did not choose \(\sqrt{5+ \sqrt x }=u\) because solving for "x" would become too labourious, though it still works.

Answer 7

I got \[\frac{ 4 }{ 3 }u^{\frac{ 3 }{ 2 }}-20u^\frac{ 1 }{ 2 }\]

Answer 8

right, then simply rewrtite the answer in terms of x and you have it

Answer 9

Cool! Thank you both.

Answer 10

I would not have chosen a sqrt of a function with sqrt for my u either there Owl Coffee

Answer 11

Thus exponential use

Answer 12

but there's more than one way to get the right answer sometimes yes

Answer 13

In my opinion, I think that changing the u to exponential form is a bit redundant. It's easier to do the arithmetic when solving for x when you see the square root sign, but that may just be me.

Answer 14

Whatever works better for you