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Special Integrals - 2

Evaluate the ...

Question

Evaluate the following integrals:

$\int \frac{x + 2}{\sqrt{x^{2} + 2 x + 3}} d x$

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Solution

$Let I = \int \frac{x + 2}{\sqrt{x^{2} + 2 x + 3}} d x We express x + 2 = A (\frac{d}{d x} (x^{2} + 2 x + 3)) + B x + 2 = A (2 x + 2) + B Equating the coefficients of x and constants, we get 1 = 2 A and 2 = 2 A + B or A = \frac{1}{2} and B = 1 ∴ I = \int \frac{\frac{1}{2} (2 x + 2) + 1}{\sqrt{x^{2} + 2 x + 3}} d x = \frac{1}{2} \int \frac{(2 x + 2)}{\sqrt{x^{2} + 2 x + 3}} d x + \int \frac{1}{\sqrt{x^{2} + 2 x + 3}} d x = \frac{1}{2} I_{1} + I_{2} . . . (1) Now, I_{1} = \int \frac{(2 x + 2)}{\sqrt{x^{2} + 2 x + 3}} d x Let x^{2} + 2 x + 3 = u On differentiating both sides, we get (2 x + 2) d x = d u ∴ I_{1} = \int \frac{1}{\sqrt{u}} d u = 2 \sqrt{u} + c_{1} = 2 \sqrt{x^{2} + 2 x + 3} + c_{1} . . . (2) And, I_{2} = \int \frac{1}{\sqrt{x^{2} + 2 x + 3}} d x = \int \frac{1}{\sqrt{x^{2} + 2 x + 1 - 1 + 3}} d x = \int \frac{1}{\sqrt{{(x + 1)}^{2} + {(\sqrt{2})}^{2}}} d x Let (x + 1) = u On differentiating both sides, we get d x = d u ∴ I_{2} = \int \frac{1}{\sqrt{{(u)}^{2} + {(\sqrt{2})}^{2}}} d u = \log |u + \sqrt{{(u)}^{2} + {(\sqrt{2})}^{2}}| + c_{2} = \log |(x + 1) + \sqrt{x^{2} + 2 x + 3}| + c_{2} . . . (3) From (1), (2) and (3), we get ∴ I = \frac{1}{2} (2 \sqrt{x^{2} + 2 x + 3} + c_{1}) + \log |(x + 1) + \sqrt{x^{2} + 2 x + 3}| + c_{2} = \sqrt{x^{2} + 2 x + 3} + \log |(x + 1) + \sqrt{x^{2} + 2 x + 3}| + c Hence, \int \frac{x + 2}{\sqrt{x^{2} + 2 x + 3}} d x = \sqrt{x^{2} + 2 x + 3} + \log |(x + 1) + \sqrt{x^{2} + 2 x + 3}| + c$

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