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$${\frac{{\mathtt{4}}{\mathtt{\,\times\,}}{pi}{\left({\mathtt{s}}\right)}}{{\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{s}}\right)}^{{\mathtt{3}}}}}{\mathtt{\,\times\,}}{\mathtt{ds}}$$

 

What is the integral of ^^?

Guest Oct 18, 2014

Best Answer 

 #1
avatar+92674 
+10

Let s+1 = u

Then s = u -1

And ds = 1 du

So we have

4pi∫(u - 1)/u3 du =

4pi [∫1/u-2 du - ∫ u-3 du ] + C =

4pi [ (-1)u-1 - (-1/2)u-2 ]  + C =

4pi [ 1/(2u2) - 1/u ] + C =

2pi [ 1- 2u]/ u2 + C =   [back substitute ....( u = s+ 1) ]

2pi [ 1 -2(s + 1) ]/ (s + 1)2 + C

2pi [ -(2s + 1) ] / (s + 1)2 + C =

-2pi (2s + 1) / (s + 1)2 + C

 

CPhill  Oct 18, 2014
 #1
avatar+92674 
+10
Best Answer

Let s+1 = u

Then s = u -1

And ds = 1 du

So we have

4pi∫(u - 1)/u3 du =

4pi [∫1/u-2 du - ∫ u-3 du ] + C =

4pi [ (-1)u-1 - (-1/2)u-2 ]  + C =

4pi [ 1/(2u2) - 1/u ] + C =

2pi [ 1- 2u]/ u2 + C =   [back substitute ....( u = s+ 1) ]

2pi [ 1 -2(s + 1) ]/ (s + 1)2 + C

2pi [ -(2s + 1) ] / (s + 1)2 + C =

-2pi (2s + 1) / (s + 1)2 + C

 

CPhill  Oct 18, 2014

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