Question-27074

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

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!