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harman billing Grade: 12
        dx/(x-1)(x^2+4(under the root))

5 years ago

Answers : (1)

Rinkoo Gupta
askIITians Faculty
80 Points
										

Integral dx/(x-1).sqrt of (x^2+4)



Put x-1=1/t



On diff w.r. to x, we get dx=-1/t^2dt



=integral -1/t^2.dt/(1/t).sqrt of ((1/t
+1)^2+4)



= -integral dt/root of (5t^2+2t+1)



=-1/root5 integral dt/sqrt of
((t+1/5)^2+(4/5)^2



=-1/root5log mod {(t+1/5) +sqrt of
(t^2+2t/5+1/5)} +C



=-1/root5 log mod [ {1/(x-1)+1/5}+sqrt of {1/(x-1)^2
+2/5.(x-1) +1/5}]+C



=-1/root5 log mod [{1/(x-1)+1/5} +
sqrt of{ (x^2+4)/5.(x-1)^2}] +C


Thanks & Regards





Rinkoo Gupta





AskIITians Faculty

3 years ago
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