How Do You Integrate f(x) = sqrt(1 + 1/x)?

[RyozKidz]

how am i going to integrate this one?

f(x) = sqrt( 1 + 1/x )

sorry for not using the symbol.. because not so familiar with that .. ^.^

 

[annoymage]

try substituting tan u = x

 

[RyozKidz]

does tan u = x work with f(x) = sqrt( 1 + (e^x)/4 ) ?

 

[annoymage]

did you solve the earlier?

does tan u = x work with f(x) = sqrt( 1 + (e^x)/4 ) ?

it didn't work

if tan u = (e^(x/2)) / 2 would work i guess

 

[annoymage]

waaaaaaa, how come, i get sqrt(4+e^x)

thats weird

 

[RyozKidz]

how did you figure out ? By intuition ?
sorry , i think i have some mistake on the earlier question i posted .

i forgot to raise the power of x to 2
the f(x) should be f(x) = sqrt( 1 + 1/(x^2) )

sorry about that ..~

 

[gabbagabbahey]

i forgot to raise the power of x to 2
the f(x) should be f(x) = sqrt( 1 + 1/(x^2) )

Have you figured this one out yet?

 

[annoymage]

owho, if your question sqrt( 1 + 1/x ) then use tan u = sqrt(x)


if sqrt( 1 + 1/(x^2) ) then use tan u = x


hmm, there are many ways, but for me, i imagine of a triangle

sqrt( 1 + 1/(x²) ) = (sqrt(x²+1))/ x

hmm how should i draw the triangle,

or wait until some other explanation people give,

i try scanning my triangle for a while. ngahahaha

 

[annoymage]

View attachment scan0001.pdf

there, hmm how do i choose those values on triangle by intuition i guess, try to make it same as your equation.

im really good in english tough, i hope someone can explain it more detail

 

[RyozKidz]

Have you figured this one out yet?

haven't ..~ no idea before annoymage replied my post~

annoymage : woww..~ never thought of that methods ...~

 

[gabbagabbahey]

In that case, you'll probably find it easiest if you make the substitution $w=\frac{1}{x}$ and integrate by parts once before making the trig substitution annoymage suggested. Give it a try and post your attempt.

 

[Dustinsfl]

I don't think integration by parts will go anywhere.

Integration tables.

You can obtain the the first integration form by substitution.

$$\int\frac{\sqrt{a+bu}}{u^2}du=\frac{-\sqrt{a+bu}}{u}+\frac{b}{2}*\int\frac{du}{u\sqrt{a+bu}}$$

$$\int\frac{du}{u\sqrt{a+bu}}=\frac{1}{\sqrt{a}}*ln{\left|\frac{\sqrt{a+bu}-\sqrt{a}}{\sqrt{a+bu}+\sqrt{a}}\right|}$$

 

[gabbagabbahey]

I don't think integration by parts will go anywhere...

$$\int\frac{\sqrt{a+bu}}{u^2}du=\frac{-\sqrt{a+bu}}{u}+\frac{b}{2}*\int\frac{du}{u\sqrt{a+bu}}$$

This is integration by parts. The tables of integrals are derived using the same methods of calculus taught to students.