Question

$\int \frac{(x^2+1)(x^2+4)}{(x^2+3)(x^2-5)}dx$

Answer 1

I = $\int \frac{(x^2+1)(x^2+4)}{(x^2+3)(x^2-5)}dx$

Since,

$$\begin{gathered} \frac{\left(x^2+1\right)\left(x^2+4\right)}{\left(x^2+3\right)\left(x^2-5\right)}=\frac{\left[\left(x^2+3\right)-2\right]\left[\left(x^2-5\right)+9\right]}{\left(x^2+3\right)\left(x^2-5\right)} \\ \Rightarrow \frac{\left(x^2+1\right)\left(x^2+4\right)}{\left(x^2+3\right)\left(x^2-5\right)}=\frac{\left(x^2+3\right)\left(x^2-5\right)+9\left(x^2+3\right)-2\left(x^2-5\right)-18}{\left(x^2+3\right)\left(x^2-5\right)} \end{gathered}$$

$\Rightarrow \frac{\left(x^2+1\right)\left(x^2+4\right)}{\left(x^2+3\right)\left(x^2-5\right)}=1+\frac{9}{\left(x^2-5\right)}-\frac{2}{\left(x^2+3\right)}-\frac{18}{\left(x^2+3\right)\left(x^2-5\right)}$ ...(i)

Let $I_1=\int \frac{1}{(x^2+3)(x^2-5)}$ and $x^2=y$

$$\begin{gathered} \Rightarrow \frac{1}{\left(y+3\right)\left(y-5\right)}=\frac{A}{\left(y+3\right)}+\frac{B}{\left(y-5\right)} \\ =\frac{A\left(y-5\right)+B\left(y+3\right)}{\left(y+3\right)\left(y-5\right)} \\ \Rightarrow \frac{1}{\left(y+3\right)\left(y-5\right)}=\frac{\left(A+B\right)y-\left(5A+3B\right)}{\left(y+3\right)\left(y-5\right)} \end{gathered}$$

Comparing coefficients, we get

$$\begin{gathered} A+B=0\mathrm{and}5A+3B=-1 \\ \mathrm{By}\mathrm{solving}\mathrm{the}\mathrm{equations},\mathrm{we}\mathrm{get} \\ A=-\frac{1}{8}\mathrm{and}B=\frac{1}{8} \end{gathered}$$

From (i), we get

$I=\int \left[1+\frac{9}{\left(x^2-5\right)}-\frac{2}{\left(x^2+3\right)}-18\left(\frac{-1}{8\left(x^2+3\right)}+\frac{1}{8\left(x^2-5\right)}\right)\right]dx$

$$\begin{gathered} \Rightarrow I=\int \left[1+\frac{27}{4\left(x^2-5\right)}+\frac{1}{\left(x^2+3\right)}\right]dx \\ \Rightarrow I=\int 1dx+\int \frac{27}{4\left(x^2-5\right)}dx+\int \frac{1}{\left(x^2+3\right)}dx \\ \therefore I=x+\frac{27}{8\sqrt{5}}\ln \left(\left|\frac{x-\sqrt{5}}{x+\sqrt{5}}\right|\right)+\frac{1}{4\sqrt{3}}{\tan }^{-1}\left(\frac{x}{\sqrt{3}}\right)+c \end{gathered}$$