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  • Integrate the functions in Exercises 1 to 24. 14

Q: Integrate the function $\frac{1}{\left(x^2+1\right)\left(x^2+4\right)}$

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Given, $\frac{1}{\left(x^2+1\right)\left(x^2+4\right)}$

Let $I=\int \frac{1}{\left(x^2+1\right)\left(x^2+4\right)}$

Now, Using partial differentiation,

$\frac{1}{\left(x^2+1\right)\left(x^2+4\right)}=\frac{A x+B}{\left(x^2+1\right)}+\frac{C x+D}{\left(x^2+4\right)}$

$\implies \frac{1}{\left(x^2+1\right)\left(x^2+4\right)}=\frac{(A x+B)\left(x^2+4\right)+(C x+D)\left(x^2+1\right)}{\left(x^2+1\right)\left(x^2+4\right)}$

$\implies 1=(A x+B)\left(x^2+4\right)+(C x+D)\left(x^2+1\right)$

$\implies 1=A x^3+4 A x+B x^2+4 B+C x^3+C x+D x^2+D$

$\implies (A+C) x^3+(B+D) x^2+(4 A+C) x+(4 B+D)=1$

Equating the coefficients of $x, x^2, x^3$ and constant value,

$A+C=0 \implies C=-A$
$B+D=0 \implies B=-D$
$4 A+C=0 \implies 4 A=-C \implies 4 A=A \implies A=0=C$
$4 B+D=1 \implies 4 B-B=1 \implies B=\frac{1}{3}=-D$

Putting these values in equation, we have

$\Rightarrow \frac{1}{\left(x^2+1\right)\left(x^2+4\right)}=\frac{(0) x+\frac{1}{3}}{\left(x^2+1\right)}+\frac{(0) x+\left(-\frac{1}{3}\right)}{\left(x^2+4\right)}$

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

$\Rightarrow \int \frac{1}{\left(x^2+1\right)\left(x^2+4\right)} d x=\frac{1}{3} \cdot \int \frac{1}{\left(x^2+1\right)} d x-\frac{1}{3} \cdot \int \frac{1}{\left(x^2+4\right)} d x$

$=\frac{1}{3} \cdot \tan ^{-1} x-\frac{1}{3} \cdot \frac{1}{2} \tan ^{-1} \frac{x}{2}+C$

$\Rightarrow I=\frac{1}{3} \tan ^{-1} x-\frac{1}{6} \tan ^{-1} \frac{x}{2}+C$

Posted by

HARSH KANKARIA

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