Question

Find m + n : $\displaystyle \sf \int \: \dfrac{ {x}^{2009} }{(1 + {x}^{2}) {}^{1106} } dx = \dfrac{1}{m} \bigg( \dfrac{ {x}^{2} }{1 + {x}^{2} } \bigg) {}^{n}$
• Class 12
• Integration ​

Answer 1

$\\ \int \frac{ {x}^{2009} }{(1 + {x}^{2} ) ^{1006} } dx = \frac{1}{m} ( \frac{ {x}^{2} }{1 + {x}^{2} } ) ^{n}$

You can easily integrate the LHS, but for now let's use RHS too.

Differentiating both the sides of the equation.

$\\ \small\frac{ {x}^{2009} }{(1 + {x}^{2}) {}^{1006} } = \frac{n}{m} \frac{(1 + {x}^{2})2 {x} - {x}^{2}(2x) }{(1 + {x}^{2} ) {}^{2} } \frac{ {x}^{2(n - 1)} }{(1 + {x}^{2} ) {}^{n - 1} }$

$\\ \small \frac{ {x}^{2009} }{(1 + {x}^{2}) {}^{1006} } = \frac{2n}{m} \frac{ {x}^{2n - 1} }{(1 + {x}^{2}) {}^{n + 1} }$

Comparing exponents and coefficient

$2n - 1 = 2009$

$\implies \boxed{n = 1005}$

$2n = m$

$\implies \boxed{ m = 2010}$

Therefore,

$\boxed{m + n = 3015}$