Problems on Techniques of Integration

We have

\begin{displaymath}\frac{x^4 + 2x^3 + 3x^2 -3 x -8}{x^2 + 2 x +5} = x^2 - 2 + \frac{x+2}{x^2 + 2 x +5}\cdot\end{displaymath}

So

\begin{displaymath}\int \frac{x^4 + 2x^3 + 3x^2 -3 x -8}{x^2 + 2 x +5}dx = \frac{x^3}{3} -2 x + \int \frac{x+2}{x^2 + 2 x +5} dx \cdot\end{displaymath}

In order to integrate $\displaystyle \frac{x+2}{x^2 + 2 x +5}$, we use the identity

\begin{displaymath}x+ 2 = \frac{1}{2} \Big(2x+2\Big) +1 \end{displaymath}

remembering the derivative of $x^2 + 2x + 5$ being $2x+2$. So

\begin{displaymath}\frac{x+2}{x^2 + 2x + 5} = \frac{1}{2} \frac{2x+2}{(x^2 + 2x + 5)} + \frac{1}{(x^2 + 2x + 5)}\cdot\end{displaymath}

This implies

\begin{displaymath}\int \frac{x+2}{x^2 + 2x + 5}dx = \frac{1}{2} \int \frac{2x+2}{x^2 + 2x + 5}dx + \int \frac{1}{x^2 + 2x + 5}dx\cdot\end{displaymath}

On one hand, we have

\begin{displaymath}\int \frac{2x+1}{x^2 + 2x + 5}dx = \ln(x^2 + 2x + 5),\end{displaymath}

and also

\begin{displaymath}\int \frac{1}{x^2 + 2x + 5}dx = \int \frac{1}{(x + 1)^2 + 4}dx = \frac{1}{2}\arctan\left(\frac{x+1}{2}\right).\end{displaymath}

We get

\begin{displaymath}\int \frac{x+2}{x^2 + 2x + 5}dx = \frac{1}{2} \ln(x^2 + 2x + 5) +\frac{1}{2}\arctan\left(\frac{x+1}{2}\right) + C.\end{displaymath}

Detailed Answer.


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