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Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
Question
Integrate the function $\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}$
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Answer

Given: $\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}$
Let $I=\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}$
Using partial fraction:
Let $\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{A x+B}{\left(x^{2}+1\right)}+\frac{C x+D}{\left(x^{2}+4\right)}$ .....(i)
$\Rightarrow \frac{1}{(x+1)\left(x^{2}+9\right)}$ = $\frac{(A x+B)\left(x^{2}+4\right)+(C x+D)\left(x^{2}+1\right)}{(x+1)\left(x^{2}+9\right)}$
$\Rightarrow 1 = (Ax + B)(x^2 + 4) + (Cx + D)(x^2 + 1)$
$\Rightarrow 1 = Ax^3 + 4Ax + Bx^2 + 4B + Cx^3 + Cx + Dx^2 + D$
$\Rightarrow 1 = (A + C)x^3 +(B + D)x^2 +(4A + C)x + (4B + D)$
Equating the coefficients of $x, x^2, x^3$ and constant value. We get:
A + C = 0 $\Rightarrow$ C = -A
B + D = 0 $\Rightarrow$ B = -D
4A + C = 0 $\Rightarrow$ 4A = -C $\Rightarrow$ 4A = A $\Rightarrow$ 3A = 0 $\Rightarrow$ A = 0 $\Rightarrow$ C = 0
4B + D = 1 $\Rightarrow$ 4B - B = 1 $\Rightarrow$ B = $\frac{1}{3}$ $\Rightarrow$ D = $\frac{-1}{3}$
Put these values in equation (i)
$\Rightarrow \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{A x+B}{\left(x^{2}+1\right)}+\frac{C x+D}{\left(x^{2}+4\right)}$
$\Rightarrow \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=$ $\frac{(0) x+\frac{1}{3}}{\left(x^{2}+1\right)}+\frac{(0) x+\left(-\frac{1}{3}\right)}{\left(x^{2}+4\right)}$
$\Rightarrow \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{\frac{1}{3}}{\left(x^{2}+1\right)}+\frac{\left(-\frac{1}{3}\right)}{\left(x^{2}+4\right)}$
$\Rightarrow \int \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x$ = $\frac{1}{3} \cdot \int \frac{1}{\left(x^{2}+1\right)} d x-\frac{1}{3} \cdot \int \frac{1}{\left(x^{2}+4\right)} d x$
$\Rightarrow \int \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x$ = $\frac{1}{3} \cdot \int \frac{1}{\left(x^{2}+1^{2}\right)} d x-\frac{1}{3} \cdot \int \frac{1}{\left(x^{2}+2^{2}\right)} d x$
= $\frac{1}{3} \cdot \tan ^{-1} x-\frac{1}{3} \cdot \frac{1}{2} \tan ^{-1} \frac{x}{2}+C$
$\Rightarrow \mathrm{I}=\frac{1}{3} \cdot \tan ^{-1} \mathrm{x}-\frac{1}{6} \tan ^{-1} \frac{\mathrm{x}}{2}+\mathrm{C}$

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