Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)} \\ \mathrm{Putting}{x}^2=\mathrm{t} \\ \mathrm{Then}, \\ \frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)}=\frac{\left(t+1\right)\left(t+2\right)}{\left(t+3\right)\left(t+4\right)}=\frac{t^2+3t+2}{t^2+7t+12} \end{gathered}$$
Degree of numerator is equal to degree of denominator.
We divide numerator by denominator.
$$\begin{gathered} 1 \\ {t}^2+7t+12\overline{)t^2+3t+2} \\ {t}^2+7t+12 \\ --- \\ \overline{-4t-10} \end{gathered}$$
$$\begin{gathered} \therefore \frac{t^2+3t+2}{t^2+7t+12}=1-\left(\frac{4t+10}{t^2+7t+12}\right) \\ \Rightarrow \frac{t^2+3t+2}{t^2+7t+12}=1-\frac{4t+10}{\left(t+3\right)\left(t+4\right)}.....\left(1\right) \\ \mathrm{Let}\frac{4t+10}{\left(t+3\right)\left(t+4\right)}=\frac{\mathrm{A}}{t+3}+\frac{\mathrm{B}}{t+4} \\ \Rightarrow \frac{4t+10}{\left(t+3\right)\left(t+4\right)}=\frac{\mathrm{A}\left(t+4\right)+\mathrm{B}\left(t+3\right)}{\left(t+3\right)\left(t+4\right)} \\ \Rightarrow 4t+10=\mathrm{A}\left(t+4\right)+\mathrm{B}\left(t+3\right) \\ \mathrm{Putting}t+4=0 \\ \Rightarrow t=-4 \\ \therefore -16+10=\mathrm{B}\left(-1\right) \\ \Rightarrow \mathrm{B}=6 \\ \mathrm{Putting}t+3=0 \\ \Rightarrow t=-3 \\ \therefore -12+10=\mathrm{A}\left(-3+4\right) \\ \Rightarrow \mathrm{A}=-2 \\ \therefore \frac{4t+10}{\left(t+3\right)\left(t+4\right)}=\frac{-2}{t+3}+\frac{6}{t+4}.....\left(2\right) \\ \mathrm{From}\left(1\right)\&\left(2\right) \\ \frac{t^2+3t+2}{t^2+7t+12}=1+\frac{2}{t+3}-\frac{6}{t+4} \\ \therefore \int \frac{\left(x^2+1\right)\left(x^2+2\right)dx}{\left(x^2+3\right)\left(x^2+4\right)}=\int dx+2\int \frac{dx}{x^2+{\left(\sqrt{3}\right)}^2}-6\int \frac{dx}{x^2+2^2} \\ =x+\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{x}{\sqrt{3}}\right)-\frac{6}{2}{\tan }^{-1}\left(\frac{x}{2}\right)+\mathrm{C} \\ =x+\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{x}{\sqrt{3}}\right)-3{\tan }^{-1}\left(\frac{x}{2}\right)+\mathrm{C} \end{gathered}$$