Question

INTEGRALS

Answer 1

K, i cant get the equation thing to work today, so

e^sqrt(t)/(sqrt(t)(e^sqrt(t)+1))dt

and i know that i need to use u substitution and i think it should be

let u= e^sqrt(t)+1 so that du= e^sqrt which gives me

(t^(1/2)(u))^-1
right?

Answer 2

\[ \frac{e^{\sqrt t}}{\sqrt t (e^{\sqrt t}+ 1)} dt \]

Answer 3

\[e ^{\sqrt(t)}/(\sqrt(t)(e^{\sqrt(t)}+1)\]

Answer 4

let sqrt (t) be a variable

Answer 5

ok..

Answer 6

let, \( e^{\sqrt t} + 1 = u \), then \( du = e^{\sqrt t} * \frac{1}{2\sqrt t} \)

Answer 7

where did 1/2sqrt(t) come from?

Answer 8

is that like chain rule?

Answer 9

chain rule ..
also i forgot to put dt at above there.

Answer 10

ok then that makes sense

Answer 11

i think you can handle the rest

Answer 12

i think so haha

Answer 13

wait actually i could use some help, i dont understand how i can use du seeing as i dont have 1/2sqrt(t) in my equation..

Answer 14

you have \( \frac{e^\sqrt t}{\sqrt t}\) in above equation .. you can balance 1/2 by multiplying both numerator and denominator with 2

Answer 15

so that will give me 2e^sqrt(t)/2sqrt(t)?

Answer 16

and if i multiply both by to it still doesnt fit the du... because then i have 2e^sqrt(t)

Answer 17

Or
let sqrt(t)=z
Differentiating both sides
dt/sqrt(t)=2dz

-----<>
2 e^z dz/(e^z+1)
let e^z+1=p
differt we get
e^z dz=dp

2dp/p

intgrating we get
2ln p+c
2ln(e^z+1)+c
2ln(e^sqrt(t)+1)+c

Answer 18

click on show steps
http://www.wolframalpha.com/input/?i=integrate+e^%28sqrt%28t%29%29%2F%28sqrt%28t%29%28e^sqrt%28t%29+%2B+1%29