Integral Challenge II: Calculate Int. 2 to 7

[lfdahl]

Calculate the definite integral\[ \int_{2}^{7} \frac{x}{1-\sqrt{2+x}}\,dx \]- without the use of an integral calculator

 

[alyafey22]

The function has a singularity at x=-1.

 

[Opalg]

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]

 

[lfdahl]

The function has a singularity at x=-1.

Thankyou, zaidalyafey, for pointing this out. I have chosen new limits for the problem.

 

[lfdahl]

[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.

 

[Opalg]

[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]

 

[lfdahl]

[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