How can i get the answer to this improper integral: $\int_{0}^{1}\frac{1}{(x-2)\sqrt{x^2+2x} } \;\mathrm{d}x $ ?
I have tried solving this in many different ways but i can't get a solution. The question is from multiple choice quiz that you dont have so much time to solve so im guessing the process shouldn't be very long, the correct answer from the quiz is supposed to be "is an improper integral, converging to $\alpha $ <0"
1 Answer
You may use the limit comparison test. Note that at $x=0$, there is division by zero, so this is an improper integral. Near $x=0$ we have
\[\frac{1}{(x-2)\sqrt{x^2+2x}}\approx \frac{1}{(0-2)\sqrt{0^2+2x}}\approx -\frac{1}{2\sqrt{2}\sqrt{x}}.\]
This means that
\[\lim_{x \rightarrow 0} \frac{\frac{1}{(x-2)\sqrt{x^2+2x}}}{-\frac{1}{2\sqrt{2}\sqrt{x}}}=1.\]
It is easy ti see that the improper integral
\[\int_0^1 -\frac{1}{2\sqrt{2}\sqrt{x}}dx\]
is convergent, and hence by the Limit Comparison Theorem $\int_0^1 \frac{1}{(x-2)\sqrt{x^2+2x}} dx$ also converges, and since on $(0,1)$ we have
\[\frac{1}{(x-2)\sqrt{x^2+2x}}<0,\]
this improper integral should converge to a negative number.
Erdos
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THANK YOU! Ive been stuck on this for approximately 2 days now, i really appreciate it :)
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