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Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals5 Marks
Question
Integrate the rational function $\frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}$
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Answer

We have, $\frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}=1-\frac{\left(4 x^{2}+10\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}$
Let $\frac{\left(4 x^{2}+10\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}=\frac{A x+B}{\left(x^{2}+3\right)}+\frac{C x+D}{\left(x^{2}+4\right)}$
$4x^2 + 10 = (Ax + B)(x^2 + 4) + (Cx + D)(x^2 + 3)$
$\Rightarrow 4x^2 + 10 = Ax^3 + 4Ax + Bx^2 + 4B + Cx^3 + 3Cx + Dx^2 + 3D$
$\Rightarrow 4x^2 + 10 = (A+C)x^3 + (B + D)x^2 + (4A + 3C)x + (4B + 3D)$
Equating the coefficients of $x^3, x^2, x$ and constant term, we get,
$A + C = 0$
$B + D = 0$
$4A + 3C = 0$
$4B + 3 B = 0$
On solving these equations, we get,
$A = 0, B = -2, C = 0$ and $D = 6$
Therefore,
$\frac{\left(4 x^{2}+10\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}=\frac{-2}{\left(x^{2}+3\right)}+\frac{6}{\left(x^{2}+4\right)}$
$\frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}=1-\left(\frac{-2}{\left(x^{2}+3\right)}+\frac{6}{\left(x^{2}+4\right)}\right)$
$\int \frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)} d x=\int\left\{1+\frac{2}{\left(x^{2}+3\right)}-\frac{6}{\left(x^{2}+4\right)}\right\} d x$
$=\left\{1+\frac{2}{x^{2}+(\sqrt{3})^{2}}-\frac{6}{\left(x^{2}+2^{2}\right)}\right\}$
$= x+2\left(\frac{1}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}\right)-6\left(\frac{1}{2} \tan ^{-1} \frac{x}{2}\right)+C$
$=x+\frac{2}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}-3 \tan ^{-1} \frac{x}{2}+C$

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