The value of 'k' is=?

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1 Answer
Feb 25, 2018

1

Explanation:

Since #x^2-2x+4 = (x-1)^2+3#, we substitute

#x-1 = sqrt{3} tan theta, qquad dx = sqrt{3} sec^2 theta #

Again, @ #x=1# we have #theta=0# and @#x=2#, we have
#theta = tan^{-1}(1/sqrt{3})=pi/6#. Thus, the integral becomes

#int_1^2 dx/(x^2-2x+4)^{3/2} = int_0^{pi/6} {sqrt{3} sec^2 theta d theta}/{3sqrt{3}sec^3 theta} = 1/3int_0^{pi/6}cos theta d theta = 1/3 sin theta|_0^{pi/6}=1/6#

Hence
#k/{k+5} = 1/6 implies k=1#