How can I go from $\int_{-\infty}^{0} \dfrac{-kx}{(x^2+d^2)^{\frac{3}{2}}}dx$ to $\dfrac{-k}{2}\int_{-\infty}^{0} \dfrac{d(x^2+d^2)}{(x^2+d^2)^{3/2}}$
This is a step from my physics book and it doesn't explain it. Can you help?
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By differentiation, $d(x^2+d^2)=2x\, dx$. Hence $kx\, dx = \frac{k}{2} \, d(x^2+d^2)$.
By the way, please don't confuse the parameter $d$ of your integral with the differential!
Siminore
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Oh I understand now. It's possible to solve the integral without doing that though, right? What if we say $x^2 + d^2 = y$? – user3601507 Jun 24 '14 at 11:55
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Yes, of course. – Siminore Jun 24 '14 at 11:56
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Assuming $d$ is a constant: $$d(x^2+d^2) = d(x^2) + d(d^2) = 2x dx + 0.$$ So $$\frac{-k}{2}\int_{-\infty}^0\frac{d(x^2+d^2)}{(x^2+d^2)^{3/2}} = \frac{-k}{2}\int_{-\infty}^0\frac{2x dx}{(x^2+d^2)^{3/2}} = \int_{-\infty}^0\frac{-kx dx}{(x^2+d^2)^{3/2}}, $$ which was to be proven.
fromGiants
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