Find the integral of $\dfrac{1}{\sqrt{x^2-4x+5}}$.

Find the integral of dfrac1√(x²-4x+5).

Question 119 of 129

\dfrac{1}{\sqrt{x^2-4x+5}}

Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2023Subjective· 3mImportance★★★★★

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Completing the square gives $(x-2)^2+1$ under the root, matching the standard integral $\int\dfrac{du}{\sqrt{u^2+a^2}}=\ln|u+\sqrt{u^2+a^2}|+c$ .

Complete the square:

$x^2-4x+5 = (x-2)^2+1$

Let $u=x-2$ , so $du=dx$ :

$\int\frac{dx}{\sqrt{x^2-4x+5}} = \int\frac{du}{\sqrt{u^2+1}}$

This matches the standard result $\int\dfrac{du}{\sqrt{u^2+a^2}} = \ln\left|u+\sqrt{u^2+a^2}\right|+c$ with $a=1$ :

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