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Question Number 27074 by Tinkutara last updated on 01/Jan/18

	Answered by mrW1 last updated on 01/Jan/18

	Commented by mrW1 last updated on 02/Jan/18
	$dm=λdx=(a_0 +b_0 x^2 )dx dF=((Gmdm)/((a+x)^2 ))=((Gm(a_0 +b_0 x^2 )dx)/((a+x)^2 )) F=∫_0 ^( L) ((Gm(a_0 +b_0 x^2 )dx)/((a+x)^2 )) F=Gm∫_0 ^( L) (((a_0 +b_0 x^2 )dx)/((a+x)^2 )) F=Gm{a_0 [(1/(a+x))]_L ^0 +b_0 ∫_0 ^( L) ((x^2 dx)/((a+x)^2 ))} F=Gm{a_0 ((1/a)−(1/(a+L)))+b_0 [x−2a ln (x+a)−(a^2 /(x+a))]_0 ^L } F=Gm{a_0 ((1/a)−(1/(a+L)))+b_0 (L+2a ln (a/(a+L))+a−(a^2 /(a+L)))} F=Gm{((a_0 L)/(a(a+L)))+b_0 [2a ln (a/(a+L))+(((2a+L)L)/(a+L))]}$
	$d m = λ d x = (a_{0} + b_{0} x^{2}) d x$ $d F = \frac{G m d m}{{(a + x)}^{2}} = \frac{G m (a_{0} + b_{0} x^{2}) d x}{{(a + x)}^{2}}$ $F = \int_{0}^{L} \frac{G m (a_{0} + b_{0} x^{2}) d x}{{(a + x)}^{2}}$ $F = G m \int_{0}^{L} \frac{(a_{0} + b_{0} x^{2}) d x}{{(a + x)}^{2}}$ $F = G m {a_{0} {[\frac{1}{a + x}]}_{L}^{0} + b_{0} \int_{0}^{L} \frac{x^{2} d x}{{(a + x)}^{2}}}$ $F = G m {a_{0} (\frac{1}{a} - \frac{1}{a + L}) + b_{0} {[x - 2 a \ln (x + a) - \frac{a^{2}}{x + a}]}_{0}^{L}}$ $F = G m {a_{0} (\frac{1}{a} - \frac{1}{a + L}) + b_{0} (L + 2 a \ln \frac{a}{a + L} + a - \frac{a^{2}}{a + L})}$ $F = G m {\frac{a_{0} L}{a (a + L)} + b_{0} [2 a \ln \frac{a}{a + L} + \frac{(2 a + L) L}{a + L}]}$
	Commented by mrW1 last updated on 02/Jan/18

	$\int \frac{x^{2}}{{(a + x)}^{2}} d x$ $= \int \frac{x^{2} + 2 a x + a^{2} - 2 a x - a^{2}}{{(a + x)}^{2}} d x$ $= \int [1 - \frac{2 a x + a^{2}}{{(a + x)}^{2}}] d x$ $= \int [1 - \frac{2 a x + 2 a^{2} - a^{2}}{{(a + x)}^{2}}] d x$ $= x - \int \frac{2 a (x + a)}{{(a + x)}^{2}} d x + a^{2} \int \frac{1}{{(a + x)}^{2}} d x$ $= x - 2 a \int \frac{1}{a + x} d x - \frac{a^{2}}{a + x}$ $= x - 2 a \ln ∣ a + x ∣ - \frac{a^{2}}{a + x} + C$
	Commented by Tinkutara last updated on 02/Jan/18

	$H o w d o y o u s o l v e d \int \frac{x^{2}}{{(a + x)}^{2}} d x ?$
	Commented by Tinkutara last updated on 02/Jan/18
	There is slight mistake in ln term. Answer given is:
	Commented by Tinkutara last updated on 02/Jan/18

	Commented by mrW1 last updated on 02/Jan/18

	$p l e a s e r e c h e c k t h e a n s w e r i n y o u r b o o k .$ $i t h i n k t h e l a s t t e r m s h o u l d b e$ $+ 2 {a b}_{0} \ln \frac{a}{a + L} o r$ $- 2 {a b}_{0} \ln \frac{a + L}{a}$
	Commented by mrW1 last updated on 02/Jan/18

	${[- 2 a \ln (a + x)]}_{0}^{L} = - 2 a [\ln (a + L) - \ln a] = - 2 a \ln \frac{a + L}{a} = 2 a \ln \frac{a}{a + L}$
	Commented by Tinkutara last updated on 03/Jan/18
	Thank you very much Sir!
	Answered by ajfour last updated on 02/Jan/18

	$F = \int_{0}^{L} \frac{G m (a_{0} + b_{0} x^{2}) d x}{{(x + a)}^{2}}$ $= G m \int_{a}^{a + L} \frac{[a_{0} + b_{0} {(t - a)}^{2}] d t}{t^{2}}$ $= G m (- \frac{a_{0}}{t}) ∣_{a}^{a + L} + b_{0} G m \int_{a}^{a + L} [1 - \frac{2 a}{t} + \frac{a^{2}}{t^{2}}] d t$ $= \frac{{G m a}_{0} L}{a (a + L)} + b_{0} G m [L - 2 a \ln (\frac{a + L}{a}) + \frac{a^{2} L}{a (a + L)}]$ $= G m [\frac{L (a_{0} + a^{2} b_{0})}{a (a + L)} + b_{0} L - 2 {a b}_{0} \ln (1 + \frac{L}{a})] .$
	Commented by mrW1 last updated on 02/Jan/18

	$t h a n k s f o r c o n f i r m i n g t h e r e s u l t!$
	Commented by Tinkutara last updated on 03/Jan/18
	Thank you very much Sir!
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