Rinkoo Gupta
Last Activity: 10 Years ago
integral a/(b+ce^x) dx
put b+ce^x=t
on diff.w.r.t. x, we get
ce^x dx=dt
integral adt/ce^x.t
=integral a dt/t(t-b)
=a/bintegral[1/(t-b)-1/t]dt
=a/b[log(mod(t-b))-log(mod(t))]+d
=a/b[log(mod(ce^x))-log(mod(b+ce^x))]+d
=a/b[cx-log(mod(b+ce^x))]+d
=-a/b[log(mod(b+ce^x))-cx]+d
comparing the result to the right hand side of the question , it is clear that K=a/b Ans.
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Rinkoo Gupta
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