int frac{x}{sqrt{x^2-2x+5}} dx= | vedclass

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Question

(A) Let $I = \int \frac{x}{\sqrt{x^2-2x+5}} dx$. We can write the numerator as $x = \frac{1}{2}(2x-2) + 1$. So,$I = \int \frac{\frac{1}{2}(2x-2)+1}{\s

Vedclass · Accepted Answer

$\sqrt{x^2-2x+5}+\ln|x-1+\sqrt{x^2-2x+5}|+c$

Vedclass · Answer

(A) Let $I = \int \frac{x}{\sqrt{x^2-2x+5}} dx$. We can write the numerator as $x = \frac{1}{2}(2x-2) + 1$. So,$I = \int \frac{\frac{1}{2}(2x-2)+1}{\sqrt{x^2-2x+5}} dx = \frac{1}{2} \int \frac{2x-2}{\sqrt{x^2-2x+5}} dx + \int \frac{1}{\sqrt{(x-1)^2+2^2}} dx$. For the first integral,let $u = x^2-2x+5$,then $du = (2x-2) dx$. $\frac{1}{2} \int u^{-1/2} du = \frac{1}{2} \cdot 2u^{1/2} = \sqrt{x^2-2x+5}$. For the second integral,use the formula $\int \frac{1}{\sqrt{t^2+a^2}} dt = \ln|t+\sqrt{t^2+a^2}|$. Here $t = x-1$ and $a = 2$,so $\int \frac{1}{\sqrt{(x-1)^2+2^2}} dx = \ln|x-1+\sqrt{(x-1)^2+4}| = \ln|x-1+\sqrt{x^2-2x+5}|$. Combining these,$I = \sqrt{x^2-2x+5} + \ln|x-1+\sqrt{x^2-2x+5}| + c$.

$\int \frac{x}{\sqrt{x^2-2x+5}} dx=$

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