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Q18  Integrate the functions \frac{5x -2 }{1+ 2x +3x^2 }

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best_answer

let 
\\5x+2 = A\frac{d}{dx}(1+2x+3x^2)+B\\ 5x+2= A(2+6x)+B = 2A+B+6Ax
By comparing the coefficients and constants we get the value of A and B

A = 5/6 and B =-11/3
 

NOW,
I = \frac{5}{6}\int \frac{6x+2}{3x^2+2x+1}dx-\frac{11}{3}\int \frac{dx}{3x^2+2x+1}
I = I_{1}-\frac{11}{3}I_{2}...........................(i)

put 3x^2+2x+1 =t \Rightarrow (6x+2)dx =dt
Thus 
I_{1}=\frac{5}{6}\int \frac{dt}{t} =\frac{5}{6}\log t =\frac{5}{6}\log (3x^2+2x+1)+c1
I_{2}= \int \frac{dx}{3x^2+2x+1} = \frac{1}{3}\int\frac{dx}{(x+1/3)^2+(\sqrt{2}/3)^2}
                                               \\=\frac{1}{\sqrt{2}}\tan^{-1}(\frac{3x+1}{\sqrt{2}})+c2

\therefore I = I_1+I_2
I = \frac{5}{6}\log(3x^2+2x+1)-\frac{11}{3}\frac{1}{\sqrt{2}}\tan^{-1}(\frac{3x+1}{\sqrt2})+C

 

Posted by

manish

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