Let #160;I= #8747; #160;x+2 #160;x2+x+1 #160;dxAlso, #160; #160;x+2= #955;ddx #160;x2+x+1+ #956; #8658;x+2= #955;2x+1+ #956; #8658;x ...

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Special Integrals - 2

∫ x+2 x 2+x+1...

Question

$\int (x + 2) \sqrt{x^{2} + x + 1} d x$

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Solution

$Let I = \int (x + 2) \sqrt{x^{2} + x + 1} d x Also, x + 2 = λ \frac{d}{d x} (x^{2} + x + 1) + μ \Rightarrow x + 2 = λ (2 x + 1) + μ \Rightarrow x + 2 = (2 λ) x + λ + μ Equating coefficient of like terms 2 λ = 1 \Rightarrow λ = \frac{1}{2} And λ + μ = 2 \Rightarrow \frac{1}{2} + μ = 2 \Rightarrow μ = \frac{3}{2} ∴ I = \int [(\frac{1}{2} (2 x + 1) + \frac{3}{2}) \sqrt{x^{2} + x + 1}] d x = \frac{1}{2} \int (2 x + 1) \sqrt{x^{2} + x + 1} d x + \frac{3}{2} \int \sqrt{x^{2} + x + 1} d x = \frac{1}{2} \int (2 x + 1) \sqrt{x^{2} + x + 1} d x + \frac{3}{2} \int \sqrt{x^{2} + x + {(\frac{1}{2})}^{2} - {(\frac{1}{2})}^{2} + 1} d x = \frac{1}{2} \int (2 x + 1) \sqrt{x^{2} + x + 1} d x + \frac{3}{2} \int \sqrt{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d x Let x^{2} + x + 1 = t \Rightarrow (2 x + 1) d x = d t Then, I = \frac{1}{2} \int \sqrt{t} d t + \frac{3}{2} \int \sqrt{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d x = \frac{1}{2} \int t^{\frac{1}{2}} d t + \frac{3}{2} [\frac{x + \frac{1}{2}}{2} \sqrt{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} + \frac{3}{8} \log |(x + \frac{1}{2}) + \sqrt{{(x + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}}|] + C = \frac{1}{2} [\frac{t \frac{3}{2}}{\frac{3}{2}}] + \frac{3}{8} (2 x + 1) \sqrt{x^{2} + x + 1} + \frac{9}{16} \log |(x + \frac{1}{2}) + \sqrt{x^{2} + x + 1}| + C = \frac{1}{3} {(x^{2} + x + 1)}^{\frac{3}{2}} + \frac{3}{8} (2 x + 1) \sqrt{x^{2} + x + 1} + \frac{9}{16} \log |(x + \frac{1}{2}) + \sqrt{x^{2} + x + 1}| + C$

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