1. ## Integrals, #1

Hi
Can you solve the problem no. 216 in the atachment ?!
i know I will need the substitution u = ln t at the first

and then ??
I wasted 2 hours ;(
then i tried k = u-1 or k = u+1 but it didnt work ..

another Question: in problem no.220 :
a = 1 , Right ?
just check it if its right or not, You are not need to solve it for me

Thank you
sorry for my bad english ;x

2. I got it !!

it will = sec^-1 [ln t + 1] + C

Solving by :
Sqrt( u . (u+2) )
= sqrt [ (u+0) . (u+1+1) ]
= sqrt [ (u+1-1) . (u+1+1) ]
then i will mulitplyed the brackets
it will be = sqrt [ (u+1)^2 - 1 ]
put k = u+1
dk = du
it will be k . Sqrt(k^2 -1)
Trigonometric substitution k = sec θ

... etc

Right ??
One question : it allowed to multiply the brackets in BLUE making u+1 as an one thing??
i hope u will understand what i mean !

$\int \frac{dt}{t(1+ln(t))\sqrt{ln(t)(2+ln(t)}}$

let u = lnt .... du = dt/t .... tdu=dt

$\int \frac{({\color{red}\rlap{/}}t)du}{{\color{red}\rlap{/}}t(1+u)\sqrt{u(2+u)}}$

$\int \frac{du}{(1+u)\sqrt{u^2+2u+1-1}}$

$\int frac {du}{(1+u)\sqrt{(u+1)^2 -1 } }$

let x = u+1 ........dx = du

$\int \frac{dx}{x\sqrt{x^2 -1 }}$

I think you can solve this right

4. thank you Amer,
i got it
funny Question!

Check a in problem 220 plz
it will be a = 1 right ??
dont solve it for me
just check it !

Thank you ^^

5. a=1 that's correct

6. Thank you.