Let #160;I= #8747; #160;x 2 #160;2x2 6x+5 #160;dxAlso, #160;x 2= #955;ddx2x2 6x+5+ #956; #8658;x 2=4 #955;x+ #956; 6 #955;Equating #16 ...

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Differentiation under Integral Sign

∫ x-22 x 2-6 ...

Question

$\int (x - 2) \sqrt{2 x^{2} - 6 x + 5} d x$

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Solution

$Let I = \int (x - 2) \sqrt{2 x^{2} - 6 x + 5} d x Also, x - 2 = λ \frac{d}{d x} (2 x^{2} - 6 x + 5) + μ \Rightarrow x - 2 = (4 λ) x + μ - 6 λ Equating the coefficient of like terms 4 λ = 1 \Rightarrow λ = \frac{1}{4} And μ - 6 λ = - 2 \Rightarrow μ - 6 \times \frac{1}{4} = - 2 \Rightarrow μ = - 2 + \frac{3}{2} = - \frac{1}{2} ∴ I = \int [\frac{1}{4} (4 x - 6) - \frac{1}{2}] \sqrt{2 x^{2} - 6 x + 5} d x = \frac{1}{4} \int (4 x - 6) \sqrt{2 x^{2} - 6 x + 5} d x - \frac{1}{2} \int \sqrt{2 x^{2} - 6 x + 5} d x Let 2 x^{2} - 6 x + 5 = t \Rightarrow (4 x - 6) d x = d t ∴ I = \frac{1}{4} \int t^{\frac{1}{2}} d t - \frac{1}{2} \int \sqrt{2 (x^{2} - 3 x + \frac{5}{2})} d x = \frac{1}{4} \int t^{\frac{1}{2}} - \frac{\sqrt{2}}{2} \int \sqrt{x^{2} - 3 x + {(\frac{3}{2})}^{2} - {(\frac{3}{2})}^{2} + \frac{5}{2}} d x = \frac{1}{4} [\frac{t^{\frac{3}{2}}}{\frac{3}{2}}] - \frac{1}{\sqrt{2}} \int \sqrt{{(x - \frac{3}{2})}^{2} - \frac{9}{4} + \frac{5}{2}} d x = \frac{1}{6} t^{\frac{3}{2}} - \frac{1}{\sqrt{2}} \int \sqrt{{(x - \frac{3}{2})}^{2} - \frac{9 + 10}{4}} d x = \frac{1}{6} t^{\frac{3}{2}} - \frac{1}{\sqrt{2}} \int \sqrt{{(x - \frac{3}{2})}^{2} + {(\frac{1}{2})}^{2}} d x = \frac{1}{6} {(2 x^{2} - 6 x + 5)}^{\frac{3}{2}} - \frac{1}{\sqrt{2}} [(\frac{x - \frac{3}{2}}{2}) \sqrt{{(x - \frac{3}{2})}^{2} + {(\frac{1}{2})}^{2}} + \frac{1}{8} \log |(x - \frac{3}{2}) + \sqrt{x^{2} - 3 x + \frac{5}{2}}|] + C = \frac{1}{6} {(2 x^{2} - 6 x + 5)}^{\frac{3}{2}} - \frac{1}{\sqrt{2}} [\frac{2 x - 3}{4} \sqrt{x^{2} - 3 x + \frac{5}{2}} + \frac{1}{8} \log |\frac{2 x - 3}{2} + \sqrt{x^{2} - 3 x + \frac{5}{2}}|] + C$

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