How to approach integral ∫ 1 ( x 2 + 7 x + 4 ) 2 d x

Question

How to approach integral   ∫      1                  (                  x                      2                          +        7        x        +        4        )                    2              d  x

sepolturamo · Accepted Answer

We have a polynomial in the integran so we should try complete the square and then partial fraction.

\begin{aligned} \int \frac{1}{(x^{2} + 7 x + 4)^{2}} d x & = \int \frac{1}{((x + 7 / 2)^{2} - 33 / 4)^{2}} d x \\ \overset{x \mapsto x + 7 / 2}{=} \int \frac{1}{(x^{2} - 33 / 4)^{2}} d x \\ = \int (\frac{4}{33 \sqrt{33} (2 x + \sqrt{33})} + \frac{4}{33 (2 x + \sqrt{33})^{2}} + \frac{4}{33 \sqrt{33} (- 2 x + \sqrt{33})} + \frac{4}{33 (- 2 x + \sqrt{33})^{2}}) d x \end{aligned}

\begin{aligned} \int \frac{1}{(x^{2} + 7 x + 4)^{2}} d x & = \int \frac{1}{((x + 7 / 2)^{2} - 33 / 4)^{2}} d x \\ \overset{x \mapsto x + 7 / 2}{=} \int \frac{1}{(x^{2} - 33 / 4)^{2}} d x \\ = \int (\frac{4}{33 \sqrt{33} (2 x + \sqrt{33})} + \frac{4}{33 (2 x + \sqrt{33})^{2}} + \frac{4}{33 \sqrt{33} (- 2 x + \sqrt{33})} + \frac{4}{33 (- 2 x + \sqrt{33})^{2}}) d x \end{aligned}

The last primitive is just using change of variables, solving and substitute back we have

\int \frac{1}{(x^{2} + 7 x + 4)^{2}} d x = \frac{1}{1089} (2 \sqrt{33} \log (7 + \sqrt{33} + 2 x) - 2 \sqrt{33} \log (- 7 + \sqrt{33} - 2 x) - \frac{33 (2 x + 7)}{x^{2} + 7 x + 4}) + C

How to approach integral ∫ 1 ( x

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