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  • Integrate the functions in Exercises 1 to 24. 14

Integrate the functions in Exercises 1 to 24.

    Q14.    \frac{1}{(x^2 + 1)(x^2 +4)}

Answers (1)

best_answer

Given,

\frac{1}{(x^2 + 1)(x^2 +4)}

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

Now, Using partial differentiation,

\frac{1}{(x^2 + 1)(x^2 +4)} = \frac{Ax + B}{(x^2 + 1)} + \frac{Cx +D}{(x^2 +4)}

\implies \frac{1}{(x^2 + 1)(x^2 +4)} = \frac{(Ax + B)(x^2 +4) + (Cx +D)(x^2 + 1)}{(x^2 + 1)(x^2 +4)}
\\ \implies1 = (Ax + B)(x^2 + 4)+(Cx + D)(x^2 + 1) \\ \implies 1 = Ax^3 +4Ax+ Bx^2 + 4B+ Cx^3 + Cx + Dx^2 + D \\ \implies (A+C)x^3 +(B+D)x^2 +(4A+C)x + (4B+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

4A + C =0 \implies 4A = -C  \implies 4A = A  \implies  A = 0 = C

4B + D = 1 \implies 4B – B = 1 \implies B = 1/3 = -D

Putting these values in equation, we have

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

Posted by

HARSH KANKARIA

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