1. ## Indefinite Integral

This is what I got
$\displaystyle \int(1+\frac{1}{x})^{-2}(\frac{1}{x^2})\,dx=-(x+1)^{-1}+C$
The book says the answer is $\displaystyle (1+\frac{1}{x})^{-1}+C$
Thanks

2. The actual problem would help immensely.

3. The actual problem was
$\displaystyle \int(1+\frac{1}{x})^{-2}(\frac{1}{x^2})\,dx$

4. They made a different substitution than you:

$\displaystyle \int (1+\frac{1}{x})^{-2} \frac{1}{x^{2}} dx$

$\displaystyle u = 1+\frac{1}{x}$
$\displaystyle du = \frac{-1}{x^{2}}dx$
$\displaystyle -\int (u)^{-2}du$

etc.

5. Originally Posted by ANDS!
They made a different substitution than you:

$\displaystyle \int (1+\frac{1}{x})^{-2} \frac{1}{x^{2}} dx$

$\displaystyle u = 1+\frac{1}{x}$
$\displaystyle du = \frac{-1}{x^{2}}dx$
$\displaystyle -\int (u)^{-2}du$

etc.
Thanks. I see how to get the answer in the book now, but I can't figure out what I did wrong to get a different answer.

6. Originally Posted by ione
Thanks. I see how to get the answer in the book now, but I can't figure out what I did wrong to get a different answer.

7. Originally Posted by ione
Thanks. I see how to get the answer in the book now, but I can't figure out what I did wrong to get a different answer.
Actually both solutions are the same.

$\displaystyle \left(1+\frac{1}{x}\right)^{-1}+C_1$

$\displaystyle =~ \frac{1}{1+\frac{1}{x}}+C_1$

$\displaystyle =~ \frac{x}{x+1}+C_1$

$\displaystyle =~ \frac{x+1-1}{x+1}+C_1$

$\displaystyle =~ 1+\frac{-1}{x+1}+C_1$

$\displaystyle =~ -\frac{1}{x+1}+C_2$

$\displaystyle =~ -(x+1)^{-1}+C_2$

8. Originally Posted by drumist
Actually both solutions are the same.

$\displaystyle \left(1+\frac{1}{x}\right)^{-1}+C_1$

$\displaystyle =~ \frac{1}{1+\frac{1}{x}}+C_1$

$\displaystyle =~ \frac{x}{x+1}+C_1$

$\displaystyle =~ \frac{x+1-1}{x+1}+C_1$

$\displaystyle =~ 1+\frac{-1}{x+1}+C_1$

$\displaystyle =~ -\frac{1}{x+1}+C_2$

$\displaystyle =~ -(x+1)^{-1}+C_2$
They are the same!!