- #1
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I am trying to work out the integral that takes the form:
[tex]I=\int \frac{x\, dx}{(a^2+x^2)^{3/2}} [/tex]
I cannot find it in a table, so I am trying By Parts.
Letting [itex]dv=x\,dx[/itex] and letting [itex]u=(a^2+x^2)^{-3/2}[/itex]
proves to be futile since I just wind up with a similar integral again, it turns into a vicious cycle
Letting [itex]u=x[/itex] and [itex](a^2+x^2)^{-3/2}\, dx[/itex] again leaves me with another integral that is not in a table, that is, I get
[tex]I=\frac{x^2}{a^2\sqrt{a^2+x^2}}-\int\frac{x\, dx}{a^2\sqrt{a^2+x^2}}[/tex]
Is there a better way? Or should I Integrate by Parts again?
This is annoying. It is just an "intermediate step" in a Griffiths E&M problem. :grumpy:
[tex]I=\int \frac{x\, dx}{(a^2+x^2)^{3/2}} [/tex]
I cannot find it in a table, so I am trying By Parts.
Letting [itex]dv=x\,dx[/itex] and letting [itex]u=(a^2+x^2)^{-3/2}[/itex]
proves to be futile since I just wind up with a similar integral again, it turns into a vicious cycle
Letting [itex]u=x[/itex] and [itex](a^2+x^2)^{-3/2}\, dx[/itex] again leaves me with another integral that is not in a table, that is, I get
[tex]I=\frac{x^2}{a^2\sqrt{a^2+x^2}}-\int\frac{x\, dx}{a^2\sqrt{a^2+x^2}}[/tex]
Is there a better way? Or should I Integrate by Parts again?
This is annoying. It is just an "intermediate step" in a Griffiths E&M problem. :grumpy: