Question

integration of 2x-1/root over (x2-2x+5)dx​

Answer 1

$\bold{\int{\dfrac{2x-1}{\sqrt{x²-2x+5}}.dx}}$

$\bold{=\int{\dfrac{2x-2+1}{\sqrt{x²-2x+5}}.dx}}$

$\bold{=\int{\dfrac{2x-2}{\sqrt{x²-2x+5}}.dx} + \int{\dfrac{1}{\sqrt{x²-2x+5}}.dx}}$

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$\sf{\red{⇒\int{\dfrac{2x-2}{\sqrt{x²-2x+5}}.dx}}}$

$\sf{x²-2x+5=t}$

$\sf{2x-2+0=\dfrac{dt}{dx}}$

$\sf{(2x-2).dx=dt}$

$\sf{\red{=\int{\dfrac{dt}{\sqrt{t}}}}}$

$\sf{\red{=\int{t^{-1/2}.dt}}}$

$\sf{\red{=\dfrac{t^{-\frac{1}{2}+1}}{-\dfrac{1}{2}+1}+c}}$

$\sf{\red{=2\sqrt{t}+c}}$

$\sf{\red{=2\sqrt{x²-2x+5}+c}}$

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$\sf{\purple{⇒\int{\dfrac{1}{\sqrt{x²-2x+5}}.dx}}}$

$\sf{\purple{=\int{\dfrac{1}{\sqrt{x²-2x+1+4}}.dx}}}$

$\sf{\purple{=\int{\dfrac{1}{\sqrt{(x-1)²+(2)²}}.dx}}}$

$\sf{x-1=t}$

$\sf{dx =dt}$

$\sf{\purple{=\int{\dfrac{1}{\sqrt{(t)²+(2)²}}.dt}}}$

$\sf{\purple{=sinh^{-1}(\dfrac{t}{2})+c}}$

$\sf{\purple{=sinh^{-1}(\dfrac{x-1}{2})+c}}$

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$\bold{=\int{\dfrac{2x-2}{\sqrt{x²-2x+5}}.dx} + \int{\dfrac{1}{\sqrt{x²-2x+5}}.dx}}$

$=$$\boxed{\bold{2\sqrt{x²-2x+5}+sinh^{-1}(\dfrac{x-1}{2})+c}}$