Question

Evaluate the integral
S(sigma) (1/√x × e^√x) dx

Answer 1

1/sqrt(x e^sqrt(x))
or
1/(sqrt(x)e^sqrt(x))

Answer 2

which function are you trying to integrate

Answer 3

\[\int\limits_{}^{} (1/\sqrt{x} \times e ^{\sqrt{x}} ) dx\]

Answer 4

im not sure which function its just a problem a have from a worksheet

Answer 5

u=sqrt(x)

Answer 6

try u substitution

Answer 7

am i substituting the exponent sqrt of the base sqrt??

Answer 8

*or

Answer 9

substitute the exponent sqrt

Answer 10

you can also substitute both if you want to

Answer 11

ill do the exponent caould it be 1/2x^-1/2 or do i have to do U^2=...

Answer 12

du = 1/2 x^(-1/2) dx works fine

Answer 13

the x^(-1/2) will cancel with base sqrt, when you convert dx to du

Answer 14

so so far its 1/2e^sqrt(u)

Answer 15

u=sqrt(x)
du = 1/2 x^(-1/2) dx
1/(sqrt(x)e^sqrt(x))dx=1/(sqrt(x)e^u)dx
1/(sqrt(x)e^u)dx = 2sqrt(x)/(sqrt(x)e^u) du

Answer 16

du = 1/2 x^(-1/2) dx
dx = 2sqrt(x)du

Answer 17

okay i understand wat i did different

Answer 18

what will i need to do next??

Answer 19

1/(sqrt(x)e^u)dx = 2sqrt(x)/(sqrt(x)e^u) du
2sqrt(x)/(sqrt(x)e^u) du = 2/e^u du
integrate this, then replace u with sqrt(x) after integrating

Answer 20

so i would need to put ln(2/e^sqrt(x)) +c

Answer 21

how do i change 2/e^sqrtx

Answer 22

where is the ln coming from?

Answer 23

just integrate 2/e^u du
=2 e^(-u) du
=-2e^(-u)+C
then replace u with sqrt(x)
=-2e^(-sqrt(x))+C

Answer 24

hmm my teacher would always use ln wen integrating

Answer 25

but i get how you do it

Answer 26

why did you make 2 negative if you only brought up e^sqrtx

Answer 27

integrate(2 e^(-u) du)=-2e^(-u)+C
the 2 becomes negative after taking the integral
integrate(e^(-x)) = -e^(-x)

Answer 28

okay thnks for your help i really appreciate it