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## What is the integration of $\int {\frac{{{x^2}}}{{\left( {{x^2} + {a^2}} \right)\left( {{x^2} + {b^2}} \right)}}} dx$ ~~~~~[NCERT Exemp. Q. 36,Page 165 ]

# Let $I = \int {\frac{{{x^2}}}{{\left( {{x^2} + {a^2}} \right)\left( {{x^2} + {b^2}} \right)}}} dx$

Now, $\frac{{{x^2}}}{{\left( {{x^2} + {a^2}} \right)\left( {{x^2} + {b^2}} \right)}}$

# [let ${x^2} = t$ ]

$ = \frac{t}{{\left( {t + {a^2}} \right)\left( {t + {b^2}} \right)}} = \frac{A}{{\left( {t + {a^2}} \right)}} + \frac{B}{{\left( {t + {b^2}} \right)}}$

#

# $t = A\left( {t + {b^2}} \right) + B\left( {t + {a^2}} \right)$

By comparing the coefficients of $t$, we get

$A + B = 1$ …….(i)

# ${b^2}A + {a^2}B = 0$ ………(ii)

# $ \Rightarrow $${b^2}(1 – B) + {a^2}B = 0$

# $ \Rightarrow $${b^2} – {b^2}B + {a^2}B = 0$

# $ \Rightarrow $${b^2} + \left( {{a^2} – {b^2}} \right)B = 0$

# $ \Rightarrow $$B = \frac{{ – {b^2}}}{{{a^2} – {b^2}}} = \frac{{{b^2}}}{{{b^2} – {a^2}}}$

From Eq. (i), $A + \frac{{{b^2}}}{{{b^2} – {a^2}}} = 1$

# $ \Rightarrow $$A = \frac{{{b^2} – {a^2} – {b^2}}}{{{b^2} – {a^2}}} = \frac{{ – {a^2}}}{{{b^2} – {a^2}}}$

# therefore, $I = \int {\frac{{ – {a^2}}}{{\left( {{b^2} – {a^2}} \right)\left( {{x^2} + {a^2}} \right)}}} dx + \int {\frac{{{b^2}}}{{{b^2} – {a^2}}}} \cdot \frac{1}{{{x^2} + {b^2}}}dx$

# $ = \frac{{ – {a^2}}}{{\left( {{b^2} – {a^2}} \right)}}\int {\frac{1}{{{x^2} + {a^2}}}} dx + \frac{{{b^2}}}{{{b^2} – {a^2}}}\int {\frac{1}{{{x^2} + {b^2}}}} dx$

# $ = \frac{{ – {a^2}}}{{{b^2} – {a^2}}} \cdot \frac{1}{a}{\tan ^{ – 1}}\frac{x}{a} + \frac{{{b^2}}}{{{b^2} – {a^2}}} \cdot \frac{1}{b}{\tan ^{ – 1}}\frac{x}{b}$

$ = \frac{1}{{{b^2} – {a^2}}}\left[ { – a\,ta{n^{ – 1}}\frac{x}{a} + b{{\tan }^{ – 1}}\frac{x}{b}} \right]$

$ = \frac{1}{{{a^2} – {b^2}}}\left[ {{{{\mathop{\rm atan}\nolimits} }^{ – 1}}\frac{x}{a} – b{{\tan }^{ – 1}}\frac{x}{b}} \right]$

Now, $\frac{{{x^2}}}{{\left( {{x^2} + {a^2}} \right)\left( {{x^2} + {b^2}} \right)}}$

$ = \frac{t}{{\left( {t + {a^2}} \right)\left( {t + {b^2}} \right)}} = \frac{A}{{\left( {t + {a^2}} \right)}} + \frac{B}{{\left( {t + {b^2}} \right)}}$

By comparing the coefficients of $t$, we get

$A + B = 1$ …….(i)

From Eq. (i), $A + \frac{{{b^2}}}{{{b^2} – {a^2}}} = 1$

$ = \frac{1}{{{b^2} – {a^2}}}\left[ { – a\,ta{n^{ – 1}}\frac{x}{a} + b{{\tan }^{ – 1}}\frac{x}{b}} \right]$

$ = \frac{1}{{{a^2} – {b^2}}}\left[ {{{{\mathop{\rm atan}\nolimits} }^{ – 1}}\frac{x}{a} – b{{\tan }^{ – 1}}\frac{x}{b}} \right]$