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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

4 years ago
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