Check my solution?

[Abhinav453]

Is it correct?

 

[stapel]

Is it correct?View attachment 36349

It looks to me like the text in the image is as follows:

[imath]\qquad e = \int \dfrac{n^2}{n^2 - 4} dn[/imath]

[imath]\qquad \textrm{for } 2[/imath]

[imath]\qquad = \int \dfrac{x^2}{n^2 - 4} dn = \int \dfrac{n^2 - 4 + 4}{n^2 - 4} dn = \int \dfrac{7^2 - 4}{x^2 - 4} dn + \int \dfrac{4}{n^2 - 4} dn[/imath]

[imath]\qquad = \int dn + 4\int \dfrac{L}{n^2 - 2^2} dn = 4 + \textrm{ [??] } \left| \dfrac{x - 2}{n + 2}\right| + C[/imath]

[imath]\qquad = 3 + \ln\left|\dfrac{7\;\,r\,\;2}{7 + 2}\right| + C[/imath]

...which can't possibly be correct. Kindly please reply with corrections. Thank you!

 

[Dr.Peterson]

I read it as

e. [imath]\int \dfrac{x^2}{x^2 - 4} dx[/imath]​

​

sol'n:​

​

[imath]\qquad = \int \dfrac{x^2}{x^2 - 4} dx = \int \dfrac{x^2 - 4 + 4}{x^2 - 4} dx = \int \dfrac{x^2 - 4}{x^2 - 4} dx + \int \dfrac{4}{x^2 - 4} dx[/imath]​

​

[imath]\qquad = \int dx + 4\int \dfrac{1}{x^2 - 2^2} dx = x + 4\cdot\frac{1}{4}\ln \left| \dfrac{x - 2}{x + 2}\right| + C[/imath]​

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[imath]\qquad = x + \ln\left|\dfrac{x - 2}{x + 2}\right| + C[/imath]​

which does make sense to me.

But, of course, you can differentiate the result and see if you get the integrand, as a check.

(Thanks, @stapel, for the preparatory work!)

 

[Steven G]

If you are unsure about the answer to an indefinite integral, then take the derivative of the answer and see if you get back the integrand.