Question

evaluate the indefinite integral

Answer 1

\[\int\limits_{}^{}( \frac{ 1-x }{ x })^2\]

Answer 2

Opening the fraction under the integral you have:

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

So your integral becomes a sum/difference of integrals (sorry, I'm too lazy to use the symbols tool):

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

Answer 3

okay, let me write this out

Answer 4

Lets simplify the integrand
\[\frac{1-x}{x}=\frac{1}{x}-1\]
So \[\left(\frac{1-x}{x} \right )^2 = \left ( \frac{1}{x}-1 \right)^2=\frac{1}{x}-\frac{2}{x}+1\]
So Now lets integrate:
\[ \int \frac{1}{x}- \int\frac{2}{x}+\int1\]

Answer 5

Integral(1/x^2) is the same as integral( x^(-2) ) and so the answer is (-1)*x^(-1).

Answer 6

Integral(1/x) = ln(x) (natural log of x)

Answer 7

\[\int \frac{1}{x}- 2\int\frac{1}{x}+\int1=\ln x -2 \ln x + x +C\]

Answer 8

@swissgirl The first integral is (1/x^2), not (1/x).

Answer 9

Ohhh shreks ... you are right

Answer 10

Yeah, it's okay, you did a better job at putting those in a nice form, I hope he gets it.

Answer 11

\[\int \frac{1}{x^2}- 2\int\frac{1}{x}+\int1=-\frac{1}{x}-2 \ln x + x +C\]

Answer 12

okay how did 1/x^2 go to -1/x

Answer 13

Ok so \[\frac{1}{x^2}=x^{-2}\]
Now integrating with powers there is a rule as follow:
\[\int x^a=(a+1)x^{a+1}\]
So our a is -2 in this case so lets use this formula
-2+1=-1
So \[ \int x^{-2}= -1x^{-1}=\frac{-1}{x}\]

Answer 14

okay thanks