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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
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Integrate : $\int \large\frac{3x+1}{(3x^2+2x+1)^2}$$dx$

\[\begin {array} {1 1} (a)\;\frac{1}{2}(3x^2+2x+1)+c \\ (b)\;(3x^2+2x+1)+c \\ (c)\;-\frac{1}{2}(3x^2+2x+1) +c \\ (d)\;None \end {array}\]

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1 Answer

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$\int \large\frac{3x+1}{(3x^2+2x+1)^2}$$dx$
=> $ 3x^2 +2x+1=t$
=> $(6x+2) dx=dt$
=> $ (3x+1)dx =\large\frac{1}{2}.$$dt$
$\large\frac{1}{2} \int \large\frac{dt}{t^2}$
=> $\large\frac{-1}{2}$$t+c$
=> $\large\frac{-1}{2} $$(3x^2+2x+1)+c$
Hence c is the correct answer.
answered Dec 19, 2013 by meena.p
 
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