91影视

The given integral is$鈭�\frac{x^2+2x+3}{x^4+4x^2+3}dx$.

Rewrite the fraction:

$\frac{x^2+2x+3}{x^4+4x^2+3}=\frac{x^2+2x+3}{(x^2+1)\left(x^2+3\right)}$

Now use the partial fraction decomposition

role="math" localid="1649508835545" $$\begin{gathered} \frac{x^2+2x+3}{(x^2+1)\left(x^2+3\right)}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+3} \\ \frac{x^2+2x+3}{(x^2+1)\left(x^2+3\right)}=\frac{\left(Ax+B\right)\left(x^2+3\right)+\left(Cx+D\right)\left(x^2+1\right)}{(x^2+1)\left(x^2+3\right)} \\ \frac{x^2+2x+3}{(x^2+1)\left(x^2+3\right)}=\frac{A{x}^3+3Ax+B{x}^2+3B+C{x}^3+Cx+D{x}^2+D}{(x^2+1)\left(x^2+3\right)} \\ \frac{x^2+2x+3}{(x^2+1)\left(x^2+3\right)}=\frac{\left(A+C\right)x^3+\left(B+D\right)x^2+\left(3A+C\right)x+\left(3B+D\right)}{(x^2+1)\left(x^2+3\right)} \\ {x}^2+2x+3=\left(A+C\right)x^3+\left(B+D\right)x^2+\left(3A+C\right)x+\left(3B+D\right) \end{gathered}$$

So, equate the coefficient:

$$\begin{gathered} A+C=0 \\ B+D=1 \\ 3A+C=2 \\ 3B+D=3 \end{gathered}$$

On solving, we get,

$A=1,B=1,C=-1,D=0$

The partial fraction can be written as follows,

$$\begin{gathered} \frac{x^2+2x+3}{(x^2+1)\left(x^2+3\right)}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+3} \\ \frac{x^2+2x+3}{(x^2+1)\left(x^2+3\right)}=\frac{\left(1\right)x+1}{x^2+1}+\frac{\left(-1\right)x+0}{x^2+3} \\ \frac{x^2+2x+3}{(x^2+1)\left(x^2+3\right)}=\frac{x+1}{x^2+1}-\frac{x}{x^2+3} \end{gathered}$$

Now we solve the integral as:

$$\begin{gathered} 鈭�\frac{x^2+2x+3}{x^4+4x^2+3}dx \\ =鈭�\left(\frac{x+1}{x^2+1}-\frac{x}{x^2+3}\right)dx \\ =鈭�\frac{x+1}{x^2+1}dx-鈭�\frac{x}{x^2+3}dx \\ =鈭�\frac{x}{x^2+1}dx+鈭�\frac{1}{x^2+1}dx-鈭�\frac{x}{x^2+3}dx \\ =\frac{1}{2}\ln \left(\left|x^2+1\right|\right)+{\tan }^{-1}\left(x\right)-\frac{1}{2}\ln \left(\left|x^2+3\right|\right)+C \end{gathered}$$

The value of integral is$\frac{1}{2}\ln \left(\left|x^2+1\right|\right)+{\tan }^{-1}\left(x\right)-\frac{1}{2}\ln \left(\left|x^2+3\right|\right)+C$.