Question

integral(x+(x)^1/2)/(x)^2

Answer 1

just split the fraction and run the powers

Answer 2

\[\int\limits_{?}^{?} (x+\sqrt{x})\div X ^{2}\]

Answer 3

so would the 1/x^2 be lnx

Answer 4

x/x^2 = 1/x , which ints up to lnx yes

what about the other one?

Answer 5

x^(1/2 - 2)

x^(1/2 - 2 + 1)
--------------
(1/2 - 2 + 1)

Answer 6

2/3x^3/2

Answer 7

might be a negative in that someplace

Answer 8

-.5 is what i see it as

Answer 9

for my final im getting ln(x) - 2/(x)^1/2

Answer 10

is this right

Answer 11

+C, but yeah, i see it as that

Answer 12

\[\int\frac{x+x^{1/2}}{x^2}dx\]
\[\int\frac{x}{x^2}+\frac{x^{1/2}}{x^2}dx\]
\[\int{x^{-1}}+x^{-3/2}dx=\ln|x|-2x^{-1/2}+C\]

Answer 13

I just took a test so I am seeing how I did by posting questions here

Answer 14

lol, ive done that in physics before

Answer 15

heres one more integral 1/x(ln(x))^2

Answer 16

its a good feeling when you see that your getting the correct answers

Answer 17

\[\int \frac{1/x}{ln^2(x)}dx\]
the top is practically the derivative of the bottom right?

let u = ln^2(x) ; du = 2ln(x)* 1/x dx

might be a suitable substitution if you go that route

Answer 18

\[ln(ln^n(x))\to\frac{n}{x~ln(x)}\]
sooo close

Answer 19

-1/ln(x)

Answer 20

\[-ln^{-1}(x)\]
\[-(-1)ln^{-2}(x)~1/x=\frac{1}{x~ln^2(x)}\]

Answer 21

i gotta go ... good luck

Answer 22

thanks