Question

Integrate the rational functions.
$\int \frac{(x^2+1)(x^2+2)}{(x^3+3)(x^2+4)}dx$

Answer 1

Here, the degree of numerator and denominator are 4. So, we convert it into simple form by putting $x^2=t$.

$\frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}=\frac{(t+1)(t+2)}{(t+3)(t+4)}=\frac{t^2+3t+2}{t^2+7t+12}$
Since, degree of numerator and denominator is same, so it can be written as
$1-\frac{4t+10}{t^2+7t+12}$
Now, $\int \frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}dx=\int 1dx-\int \frac{(4t+10)}{t^2+7t+12}dx$
$=\int 1dx-\int \frac{4t+10}{(t+3)(t+4)}dx.....\left(i\right)$
Let $\frac{4t+10}{(t+4)(t+3)}=\frac{A}{(t+4)}+\frac{B}{(t+3)}\Rightarrow \frac{4t+10}{(t+3)(t+4)}=\frac{A(t+3)+B(t+4)}{(t+4)(t+3)}$
$\Rightarrow 4t+10=At+3A+Bt+4B\Rightarrow 4t+10=t(A+B)+(3A+4B)$
On comparing the coefficients of t and constant term on both sides, we get
$A+B=4\Rightarrow 3A+3B=12......\left(ii\right)$
and 3A+4B =10
On subtracting Eq. (iii)from Eq. (ii), we get
$-B=2\Rightarrow B=-2\text{and}A=6\therefore \frac{4t+10}{(t+4)(t+3)}=\frac{6}{t+4}-\frac{2}{t+3}$
On putting this value in Eq. (i)we get
$\int \frac{(x^2+1)(x^2+2)}{(x^3+3)(x^2+4)}dx=\int 1dx-\int [\frac{6}{t+4}-\frac{2}{t+3}]dx=\int 1dx-\int (\frac{6}{x^2+4}-\frac{2}{(x^2+3)})dx\text{[Put t=}{x}^2\text{]}\Rightarrow \int \frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}dx=\int 1dx-\int (\frac{6}{x^2+2^2}-\frac{2}{x^2+(\sqrt{3}{)}^2})dx\Rightarrow \int \frac{(x^2+1)(x^2+2)}{(x^2+3)(x^2+4)}dx=x-6\left(\frac{1}{2}ta{n}^{-1}\frac{x}{2}\right)+2\left(\frac{1}{\sqrt{3}}ta{n}^{-1}\frac{x}{\sqrt{3}}\right)+C(\because \int \frac{1}{a^2+x^2}dx=\frac{1}{a}ta{n}^{-1}\frac{x}{a})=x-3ta{n}^{-1}\frac{x}{2}+\frac{2}{\sqrt{3}}ta{n}^{-1}\frac{x}{\sqrt{3}}+C$