# Integrate this one?

1. Apr 14, 2010

### 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 .. ^.^

2. Apr 14, 2010

### annoymage

try substituting tan u = x

3. Apr 14, 2010

### RyozKidz

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

4. Apr 14, 2010

### annoymage

did you solve the earlier?

it didn't work

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

5. Apr 14, 2010

### annoymage

waaaaaaa, how come, i get sqrt(4+ex)

thats wierd

Last edited: Apr 14, 2010
6. Apr 14, 2010

### 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) )

7. Apr 14, 2010

### gabbagabbahey

Have you figured this one out yet?

8. Apr 14, 2010

### 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

9. Apr 14, 2010

### 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

10. Apr 14, 2010

### RyozKidz

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

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

Last edited: Apr 14, 2010
11. Apr 14, 2010

### 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.

12. Apr 15, 2010

### Dustinsfl

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

Integration tables.

You can obtain the the first integration form by substitution.

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

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

13. Apr 16, 2010

### gabbagabbahey

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