MHB Integral Challenge II: Calculate Int. 2 to 7

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The discussion focuses on calculating the definite integral of the function \(\int_{2}^{7} \frac{x}{1-\sqrt{2+x}}\,dx\) and highlights its improper nature due to a singularity at \(x=-1\). It is established that the integral converges only if both \(\int_{-2}^{-1} \frac{x}{1-\sqrt{2+x}}\,dx\) and \(\int_{-1}^{2} \frac{x}{1-\sqrt{2+x}}\,dx\) exist. The substitution \(y = \sqrt{2+x}\) is used to evaluate the second integral, leading to a divergence due to the logarithmic term as \(y\) approaches 1. Ultimately, the integral does not converge, confirming that it does not exist. The discussion concludes with acknowledgments to contributors for their insights on the singularity issue.
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Calculate the definite integral\[ \int_{2}^{7} \frac{x}{1-\sqrt{2+x}}\,dx \]- without the use of an integral calculator
 
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The function has a singularity at x=-1.
 
lfdahl said:
Calculate the definite integral\[ \int_{-2}^{2} \frac{x}{1-\sqrt{2+x}}\,dx \]- without the use of an integral calculator
[sp]The function has a singularity at $x=-1$, where the denominator $1-\sqrt{2+x}$ is zero. So the integral is improper, and it will only converge if each of the two integrals $$\int_{-2}^{-1} \frac{x}{1-\sqrt{2+x}}\,dx$$ and $$\int_{-1}^{2} \frac{x}{1-\sqrt{2+x}}\,dx$$ exists.

Taking the second of these integrals, $$\int_{-1}^{2} \frac{x}{1-\sqrt{2+x}}\,dx$$, make the substitution $y = \sqrt{2+x}.$ Then $2+x = y^2$, $dx = 2y\,dy$ and the integral becomes $$\int_1^2 \frac{2y(y^2-2)}{1-y}dy.$$ Since $2y(y^2-2) = (y-1)(2y^2 + 2y - 2) -2,$ the integral is equal to $$\int_1^2\bigl(-2y^2 -2y + 2 - \frac2{1-y}\bigr)dy.$$ The integral of $-2y^2 - 2y + 2$ is elementary. But the integral of $- \frac2{1-y}$ is $2\ln|1-y|$, which diverges as $y \searrow1.$

So the integral does not converge. (More informally, the integral does not exist.)[/sp]
 
ZaidAlyafey said:
The function has a singularity at x=-1.

Thankyou, zaidalyafey, for pointing this out. I have chosen new limits for the problem.
 
Opalg said:
[sp]The function has a singularity at $x=-1$, where the denominator $1-\sqrt{2+x}$ is zero. So the integral is improper, and it will only converge if each of the two integrals $$\int_{-2}^{-1} \frac{x}{1-\sqrt{2+x}}\,dx$$ and $$\int_{-1}^{2} \frac{x}{1-\sqrt{2+x}}\,dx$$ exists.

Taking the second of these integrals, $$\int_{-1}^{2} \frac{x}{1-\sqrt{2+x}}\,dx$$, make the substitution $y = \sqrt{2+x}.$ Then $2+x = y^2$, $dx = 2y\,dy$ and the integral becomes $$\int_1^2 \frac{2y(y^2-2)}{1-y}dy.$$ Since $2y(y^2-2) = (y-1)(2y^2 + 2y - 2) -2,$ the integral is equal to $$\int_1^2\bigl(-2y^2 -2y + 2 - \frac2{1-y}\bigr)dy.$$ The integral of $-2y^2 - 2y + 2$ is elementary. But the integral of $- \frac2{1-y}$ is $2\ln|1-y|$, which diverges as $y \searrow1.$

So the integral does not converge. (More informally, the integral does not exist.)[/sp]

Thankyou, Opalg, for your clearcut contribution. I am very sorry, that I was not aware of this.
Please cf. also my comment at zaidalyafey. Hopefully, the new limits are OK.
 
[sp]With the new limits, $y=2$ when $x=2$ and $y=3$ when $x=9$ (see my comment #3 above), and the integral becomes $$\int_2^3\bigl(-2y^2 -2y + 2 + \frac2{y-1}\bigr)dy = \bigl[-\tfrac23y^3 - y^2 + 2y + 2\ln(y-1)\bigr]_2^3 = -\tfrac{47}3 + 2\ln2.$$[/sp]
 
Opalg said:
[sp]With the new limits, $y=2$ when $x=2$ and $y=3$ when $x=9$ (see my comment #3 above), and the integral becomes $$\int_2^3\bigl(-2y^2 -2y + 2 + \frac2{y-1}\bigr)dy = \bigl[-\tfrac23y^3 - y^2 + 2y + 2\ln(y-1)\bigr]_2^3 = -\tfrac{47}3 + 2\ln2.$$[/sp]

Thankyou very much, Opalg, for your correct answer!
I also want to thank Zaidalyafey and Opalg for detecting the troublesome singularity :cool:
 

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