Question

integrate: from 0 to 1

(x-1) / (x+1) = the first step for the solutions says
((x+1) (x+1)) - (2 / (x+1))

Answer 1

aaaak that is ((x+1)/(x+1))

Answer 2

\[\int\limits_{0}^{1} (x-1) / (x+1)\]

then the next step says: = ((x+1) / (x+1)) - (2/(x+1)

= 1 - (2/(x+1) = x-2(ln(x+1).... the only part i get is the last step for integrating the 1 and the 1/x...

Answer 3

\[\int\limits_{0}^{1} \frac{x-1}{x+1}dx\]
\[=\int\limits\limits_{0}^{1} \frac{x-1+(1-1)}{x+1}dx\]
You add 1-1 (which is 0) on the top. We added a 0in this way because it doesn't change the expression but it helps us simplify as follows.
\[=\int\limits\limits_{0}^{1} \frac{x+1-2}{x+1}dx\]
Rearrange the top to look like this

Then separate
\[=\int\limits\limits\limits_{0}^{1} (\frac{x+1}{x+1}-\frac{2}{x+1})dx\]
Now
\[=\int\limits\limits\limits_{0}^{1} (1-\frac{2}{x+1})dx\]

Do you understand where to go from here?

Answer 4

ok i thought they were multiplying something not just adding and then subtracting one...that sure seems like a trick question.....thanks

Answer 5

it does seem like a trick question but as you get to do more and more integral problems, you'll find that this "trick" comes in handy.

basically, this is what happened...