Question

integrate 1/9+sqrt(2x) I'm supposed to solve it using u-substitution but im stumped

Answer 1

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{1}{9+\sqrt{2x}}~dx}\)

Answer 2

yes

Answer 3

I dare you to try subbing the whole bottom as u

Answer 4

that is du = 1/sqrt(2x) dx but what do i do about the sqrt(2x) ?

Answer 5

\[u=9+\sqrt{2x} \\ u-9=\sqrt{2x}\]

Answer 6

\[\int\limits \frac{u-9}{u} du\]

Answer 7

oh ok that makes alot of sense

Answer 8

thank you

Answer 9

x - 9(ln|9 + sqrt(2x)|) + C

Answer 10

just in case...
\[u=9+\sqrt{2x} \\ du=\frac{1}{\sqrt{2x}} dx \\ \text{ replace } \sqrt{2x} \text{with } u-9 \\ du=\frac{1}{u-9} dx \\ (u-9) du =dx\]
ok now checking your final answer...

Answer 11

why did you replace the first u with x and not 9+sqrt(2x)?

Answer 12

should have u-9ln|u|+C
where u=9+sqrt(2x)

Answer 13

that first u you put x?

Answer 14

can't you split it up into u/u - 9/u ?

Answer 15

you can

Answer 16

but you are integrating with respect to u
not one x and the other u

Answer 17

oh ok so the 1 would integrate back into u which i would then replace

Answer 18

right because we are integrating with respect to u
because of the du part

Answer 19

\[\int\limits (1-\frac{9}{u} ) du\\ u-9 \ln|u|+C\]

Answer 20

and final step is to replace any u you see with 9+sqrt(2x)

Answer 21

ok thank you, I'll practice more problems like this one

Answer 22

k

Answer 23

I think there is another way without u sub to do this one
would you like to see?

Answer 24

yes

Answer 25

please

Answer 26

\[\frac{1}{9+\sqrt{2x}} \cdot 1 \\=\frac{1}{9+\sqrt{2x}} \cdot \frac{9-\sqrt{2x}}{9-\sqrt{2x}} \\ =\frac{9-\sqrt{2x}}{81-2x}\]
oops spoke to soon.

Answer 27

i actually tried doing it that way but i quickly figured out i would involve multiple substitutions which idk how to do yet

Answer 28

\[\frac{1}{9+\sqrt{2x}} =\frac{1}{\sqrt{2x}} \cdot \frac{\sqrt{2x}}{9+\sqrt{2x}} \\ =\frac{1}{\sqrt{2x}} \cdot \frac{\sqrt{2x}+9-9}{9+\sqrt{2x}} \\ =\frac{1}{\sqrt{2x}} \cdot (\frac{\sqrt{2x}+9}{9+\sqrt{2x}}-\frac{9}{9+\sqrt{2x}}) \\ =\frac{1}{\sqrt{2x}} (1-\frac{9}{9+\sqrt{2x}}) \\ =\frac{1}{\sqrt{2x}}- \frac{1}{\sqrt{2x}} \cdot \frac{9}{9+\sqrt{2x}}\]
yea you still might prefer a substitution on the second one

Answer 29

so yep I spoke way too soon earlier
my bad

Answer 30

hahaha no problem but seriously thank you for the help

Answer 31

np