Integrate: 1/(x-sqrt(x+2) dx
I came up with:
(2/3)(2*ln((sqrt(x+2))-2)+ln((sqrt(x+2))-1))
but it keeps coming back the wrong answer even though I integrated correctly. Is there a way to simplify this answer, and if so, how?
I found:
Ln[x-sqrt(x+2)] +
1/3Ln[(sqrt(x+2)-2)/(sqrt(x+2) + 1)]
ok but is there a way to further simplify this answer?
To check if the two expressions are equivalent or can be simplified further, let's simplify both expressions separately.
Your expression:
(2/3)(2*ln((sqrt(x+2))-2)+ln((sqrt(x+2))-1))
Let's simplify the first part:
2*ln((sqrt(x+2))-2)
= ln((sqrt(x+2))-2)^2
= ln(x+2-4sqrt(x+2)+4)
= ln(x-4sqrt(x+2)+6)
Now let's simplify the second part:
ln((sqrt(x+2))-1)
= ln((sqrt(x+2))-1)^2
= ln(x+2-2sqrt(x+2)+1)
= ln(x-2sqrt(x+2)+3)
So, the simplified expression becomes:
(2/3)(ln(x-4sqrt(x+2)+6) + ln(x-2sqrt(x+2)+3))
Now let's simplify the alternative expression you found:
Ln[x-sqrt(x+2)] + 1/3Ln[(sqrt(x+2)-2)/(sqrt(x+2) + 1)]
To simplify, start by simplifying the logarithmic terms:
Ln[x-sqrt(x+2)] = ln(x-√(x+2))
Next, simplify the fraction inside the logarithm:
[(sqrt(x+2)-2)/(sqrt(x+2) + 1)] = [(√(x+2)-2)/(√(x+2) + 1)]
So, the simplified alternative expression becomes:
ln(x-√(x+2)) + 1/3ln[(√(x+2)-2)/(√(x+2) + 1)]
Comparing the two expressions, it appears that they are not the same. Therefore, it seems that the expression you found:
Ln[x-sqrt(x+2)] + 1/3Ln[(sqrt(x+2)-2)/(sqrt(x+2) + 1)]
is the simplest form of the integral.