$\int \frac{3 x + 1}{{({3 x}^{2} + 2 x + 1)}^{3}} d x$

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Solution

Let, $3 x^{2} + 2 x + 1 = t$

=> $(6 x + 2) d x = d t$

=> $2 (3 x + 1) d x = d t$

So, given equation becomes,

=> $\int \frac{d t}{2 t^{3}}$

=> $- \frac{1}{8 t^{4}} + C$

=> $\frac{- 1}{8 (3 x^{2} + 2 x + 1)^{4}} + C$

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