Question
Solve this : $$\int\limits_0^1 {\frac{1}{{\left( {{x^2} + 16} \right)\left( {{x^2} + 25} \right)}}} dx = ?$$
A.
$$\frac{1}{5}\left[ {\frac{1}{4}{{\tan }^{ - 1}}\left( {\frac{1}{4}} \right) - \frac{1}{5}{{\tan }^{ - 1}}\left( {\frac{1}{5}} \right)} \right]$$
B.
$$\frac{1}{9}\left[ {\frac{1}{4}{{\tan }^{ - 1}}\left( {\frac{1}{4}} \right) - \frac{1}{5}{{\tan }^{ - 1}}\left( {\frac{1}{5}} \right)} \right]$$
C.
$$\frac{1}{4}\left[ {\frac{1}{4}{{\tan }^{ - 1}}\left( {\frac{1}{4}} \right) - \frac{1}{5}{{\tan }^{ - 1}}\left( {\frac{1}{5}} \right)} \right]$$
D.
$$\frac{1}{9}\left[ {\frac{1}{5}{{\tan }^{ - 1}}\left( {\frac{1}{4}} \right) - \frac{1}{5}{{\tan }^{ - 1}}\left( {\frac{1}{5}} \right)} \right]$$
Answer :
$$\frac{1}{9}\left[ {\frac{1}{4}{{\tan }^{ - 1}}\left( {\frac{1}{4}} \right) - \frac{1}{5}{{\tan }^{ - 1}}\left( {\frac{1}{5}} \right)} \right]$$
Solution :
$$\eqalign{
& {\text{Let }}I = \int\limits_0^1 {\frac{{dx}}{{\left( {{x^2} + 16} \right)\left( {{x^2} + 25} \right)}}} \cr
& = \frac{1}{9}\int\limits_0^1 {\left( {\frac{1}{{{x^2} + 16}} - \frac{1}{{{x^2} + 25}}} \right)} dx \cr
& = \frac{1}{9}\left( {\frac{1}{4}{{\tan }^{ - 1}}\frac{x}{4} - \frac{1}{5}{{\tan }^{ - 1}}\frac{x}{5}} \right)_0^1 \cr
& = \frac{1}{9}\left[ {\frac{1}{4}{{\tan }^{ - 1}}\frac{1}{4} - \frac{1}{5}{{\tan }^{ - 1}}\frac{1}{5}} \right] \cr} $$