Let #160;I= #8747;x #160;dxx2+6x+10x=A #160;ddx #160;x2+6x+10+Bx=A #160;2x+6+Bx=2A #160;x+6A+BEquating #160;Coefficients #160;of #160;like ...

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Special Integrals - 2

∫ x x 2+6 x+1...

Question

$\int \frac{x}{\sqrt{x^{2} + 6 x + 10}} d x$

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Solution

$Let I = \int \frac{x d x}{\sqrt{x^{2} + 6 x + 10}} x = A \frac{d}{d x} (x^{2} + 6 x + 10) + B x = A (2 x + 6) + B x = (2 A) x + 6 A + B Equating Coefficients of like terms 2 A = 1 A = \frac{1}{2} 6 A + B = 0 6 \times \frac{1}{2} + B = 0 B = - 3 I = \int \frac{x d x}{\sqrt{x^{2} + 6 x + 10}} = \int (\frac{\frac{1}{2} (2 x + 6) - 3}{\sqrt{x^{2} + 6 x + 10}}) d x = \frac{1}{2} \int \frac{(2 x + 6) d x}{\sqrt{x^{2} + 6 x + 10}} - 3 \int \frac{d x}{\sqrt{x^{2} + 6 x + 3^{2} - 3^{2} + 10}} = \frac{1}{2} \int \frac{(2 x + 6) d x}{\sqrt{x^{2} + 6 x + 10}} - 3 \int \frac{d x}{\sqrt{{(x + 3)}^{2} + 1^{2}}} let x^{2} + 6 x + 10 = t \Rightarrow (2 x + 6) d x = d t I = \frac{1}{2} \int \frac{d t}{\sqrt{t}} - 3 \int \frac{d x}{\sqrt{{(x + 3)}^{2} + 1}} = \frac{1}{2} \times 2 \sqrt{t} - 3 \log |x + 3 + \sqrt{{(x + 3)}^{2} + 1}| + C = \sqrt{t} - 3 \log |x + 3 + \sqrt{x^{2} + 6 x + 10}| + C = \sqrt{x^{2} + 6 x + 10} - 3 \log |x + 3 + \sqrt{x^{2} + 6 x + 10}| + C$

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