# ∫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

Vinod Ramakrishnan Eswaran
41 Points
4 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