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∫dx/(x^2-2x+4)^3/2 please provide me detailed solution of this step wise solution . it defenite integral where lower limit is 1 and upper limit is 2

∫dx/(x^2-2x+4)^3/2          please provide me detailed solution of this step wise solution  . it defenite integral where lower limit is 1 and upper limit is 2
 
 

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Vinod Ramakrishnan Eswaran
41 Points
5 years ago
 I=\int_{1}^{2}dx/(x^{2}-2x+4)^{3/2}
I=\int_{1}^{2}dx/(x^{2}-2x+4)^{3/2}\; =>\int_{1}^{2}dx/((x-1)^{2}+3)^{3/2}
 
Substitute\: x-1\: as\: t,\: we\: get\: dx=dt
I=\int_{0}^{1}dt/(t^{2}+3)^{3/2}
We\: can\: use\: formula\: I=\int dx/(a^{2}+x^{2})^{3/2}=x/((a^{2})(a^{2}+x^{2})^{1/2}) * 
We\: get\: I=t/(3(3+t^{2})^{1/2})^{1}_{0}\:
Solving we get I=1/6
Hence this is the solution
For the proof for the * we need to substitute x as atanx and then integrate

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