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Special Integrals - 2

Evaluate the ...

Question

Evaluate the following integrals:

$\int \frac{5 x - 2}{1 + 2 x + 3 x^{2}} d x$

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Solution

$Let I = \int \frac{5 x - 2}{1 + 2 x + 3 x^{2}} d x = \int \frac{5 x - 2}{3 x^{2} + 2 x + 1} d x We express 5 x - 2 = A (\frac{d}{d x} (3 x^{2} + 2 x + 1)) + B 5 x - 2 = A (6 x + 2) + B Equating the coefficients of x and constants, we get 5 = 6 A and - 2 = 2 A + B or A = \frac{5}{6} and B = - \frac{11}{3} ∴ I = \int \frac{\frac{5}{6} (6 x + 2) - \frac{11}{3}}{3 x^{2} + 2 x + 1} d x = \frac{5}{6} \int \frac{(6 x + 2)}{3 x^{2} + 2 x + 1} d x - \frac{11}{3} \int \frac{1}{3 x^{2} + 2 x + 1} d x = \frac{5}{6} I_{1} - \frac{11}{3} I_{2} . . . (1) Now, I_{1} = \int \frac{(6 x + 2)}{3 x^{2} + 2 x + 1} d x Let 3 x^{2} + 2 x + 1 = t On differentiating both sides, we get (6 x + 2) d x = d t ∴ I_{1} = \int \frac{1}{t} d t = \log |t| + c_{1} = \log |3 x^{2} + 2 x + 1| + c_{1} . . . (2) And, I_{2} = \int \frac{1}{3 x^{2} + 2 x + 1} d x = \frac{1}{3} \int \frac{1}{x^{2} + \frac{2}{3} x + \frac{1}{3}} d x = \frac{1}{3} \int \frac{1}{x^{2} + \frac{2}{3} x + \frac{1}{9} - \frac{1}{9} + \frac{1}{3}} d x = \frac{1}{3} \int \frac{1}{{(x + \frac{1}{3})}^{2} + {(\frac{\sqrt{2}}{3})}^{2}} d x Let x + \frac{1}{3} = t On differentiating both sides, we get d x = d t ∴ I_{2} = \frac{1}{3} \int \frac{1}{{(t)}^{2} + {(\frac{\sqrt{2}}{3})}^{2}} d t = \frac{1}{3} \times \frac{1}{\frac{\sqrt{2}}{3}} \tan^{- 1} \frac{3 t}{\sqrt{2}} + c_{2} = \frac{1}{\sqrt{2}} \tan^{- 1} \frac{3 (x + \frac{1}{3})}{\sqrt{2}} + c_{2} = \frac{1}{\sqrt{2}} \tan^{- 1} \frac{3 x + 1}{\sqrt{2}} + c_{2} . . . (3) From (1), (2) and (3), we get ∴ I = \frac{5}{6} (\log |3 x^{2} + 2 x + 1| + c_{1}) - \frac{11}{3} (\frac{1}{\sqrt{2}} \tan^{- 1} \frac{3 x + 1}{\sqrt{2}} + c_{2}) = \frac{5}{6} (\log |3 x^{2} + 2 x + 1|) - \frac{11}{3} (\frac{1}{\sqrt{2}} \tan^{- 1} \frac{3 x + 1}{\sqrt{2}}) + c Hence, \int \frac{5 x - 2}{1 + 2 x + 3 x^{2}} d x = \frac{5}{6} (\log |3 x^{2} + 2 x + 1|) - \frac{11}{3} (\frac{1}{\sqrt{2}} \tan^{- 1} \frac{3 x + 1}{\sqrt{2}}) + c$

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