Question

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

Answer 1

So,m+n=1005+2010

=3015

DM me ,if u have any doubt in my solution..

Answer 2

Correct question :

$\displaystyle \sf \int \: \dfrac{ {x}^{2009} }{(1 + {x}^{2}) {}^{1006} } dx = \dfrac{1}{m} \bigg( \dfrac{ {x}^{2} }{1 + {x}^{2} } \bigg) {}^{n} + c$

Solution :

$\sf \rightarrow\displaystyle \sf \int \: \dfrac{ {x}^{2009} }{(1 + {x}^{2}) {}^{1006} } dx = \displaystyle \sf \int \dfrac{ {x}^{2009} }{ {x}^{2012} (1 + \frac{1}{ {x}^{2} } ) {}^{1006} } dx \\ \\ \\ \rightarrow \displaystyle \sf \int \dfrac{ 1}{ {x}^{3} (1 + \frac{1}{ {x}^{2} } ) {}^{1006} } dx$

$\rightarrow \sf let \: \: \: 1 + \dfrac{1}{ {x}^{2}} = t \\ \\\rightarrow \sf \frac{ - 2}{ {x}^{3} } dx=dt \\ \\ \boxed{ \rightarrow \sf \frac{ 1}{ {x}^{3} } dx= \dfrac{ - 1}{2} dt} \\ \\ \\ \rightarrow \displaystyle \sf \int \dfrac{ - dt}{ 2(t) {}^{1006} } \\ \\ \\ \rightarrow - \displaystyle \sf \int \dfrac{ (t) {}^{ - 1006} dt}{ 2} \\ \\ \\ \rightarrow - \sf \dfrac{ (t) {}^{ - 1005} }{ 2 \times ( - 1005)} + c\\ \\ \rightarrow \sf \dfrac{ 1 }{ 2010 \times (t) {}^{ 1005}} + c \\ \\ \\ \sf now \: put \: t \: = 1 + \frac{1}{ {x}^{2}} \\ \\ \rightarrow \sf \dfrac{ 1 }{ 2010 \times (1 + \frac{1}{ {x}^{2}}) {}^{ 1005}} + c \\ \\ \rightarrow \sf \frac{1}{201 0} \times \dfrac{ 1 }{ ( \frac{ {x}^{2} + 1 }{ {x}^{2}}) {}^{ 1005}} + c \\ \\ \\ \rightarrow \sf \frac{1}{201 0} \times \bigg( {\frac{ {x}^{2} }{ {x}^{2} + 1 } } \bigg)^{1005} + c$

Therefore , m = 2010 and n = 1005

=> m+n = 2010+1005

=> m+n = 3015