Let I = ∫√(5 2x+x^2)dx=∫√(x^2 2x+1+4)dx = ∫√((x 1)^2+(2)^2)dx=∫√((2)^2+(x 1)^2dx) =x 1/2√(2^2+(x 1)^2)+2 log|(x 1)+√(2^2+(x 1)^2)|+C =x 1/2√(5 2x+x^2)+2 log| ...

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Byju's Answer

Standard XII

Mathematics

Properties of Set Operation

∫√5-2 x+x2 d ...

Question

$\int \sqrt{5 - 2 x + x^{2}} d x$

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Solution

Let $I = \int \sqrt{5 - 2 x + x^{2}} d x = \int \sqrt{x^{2} - 2 x + 1 + 4} d x = \int \sqrt{(x - 1)^{2} + (2)^{2}} d x = \int \sqrt{(2)^{2} + (x - 1)^{2} d x} = \frac{x - 1}{2} \sqrt{2^{2} + (x - 1)^{2}} + 2 l o g | (x - 1) + \sqrt{2^{2} + (x - 1)^{2}} | + C = \frac{x - 1}{2} \sqrt{5 - 2 x + x^{2}} + 2 l o g | (x - 1) + \sqrt{5 - 2 x + x^{2}} | + C$

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∫√5−2x+x2dx

Let I=∫√5−2x+x2dx=∫√x2−2x+1+4dx=∫√(x−1)2+(2)2dx=∫√(2)2+(x−1)2dx=x−12√22+(x−1)2+2 log|(x−1)+√22+(x−1)2|+C=x−12√5−2x+x2+2 log|(x−1)+√5−2x+x2|+C

$\int \sqrt{5 - 2 x + x^{2}} d x$