Question

∫frac{x^{2} } / [ {(x^{2} } +4)({x^{2} +9)] dx

Answer 1

Correct Question :

$∫\frac{ {x}^{2} }{ ({x}^{2} + 4 )( {x}^{2} + 9)}</u><u>dx</u><u>$

Solution:

Put

${x}^{2} = y$

$\frac{ {x}^{2} }{( {x}^{2} + 4)( {x}^{2} + 9)} = \frac{y}{(y + 4)(y + 9)} ........(1)$

Let

$\frac{y}{(y + 4)(y + 9)} = \frac{A}{(y + 4)} + \frac{B}{y + 9} ........(2)$

Multiplying by (y+4)(y+9)

y ≡ A(y+9)+B(y+4)

Putting y= -4

=> -4= A(-4+9)

=> A=-4/5

Putting y =-9

=> -9= B (-9+4)

=>B=9/5

From (2)

$\frac{y}{(y + 4)(y + 9)} = \frac{ \frac{ - 4}{5} }{y + 4} + \frac{ \frac{9}{5} }{y + 9} \\ \\ = > \frac{ {x}^{2} }{( {x}^{2} + 4)( {x}^{2} + 9)} = \frac{ \frac{ - 4}{5} }{ {x}^{2} + 4 } + \frac{ \frac{9}{5} }{ {x}^{2} + 9}$

Therefore

$∫ \frac{ {x}^{2} }{( {x}^{2} + 4)( {x}^{2} + 9) } dx \\ \\ = \frac{ - 4}{5} ∫ \frac{1}{ {2}^{2} + {x}^{2} } + \frac{9}{5} ∫ \frac{1}{ {3}^{2} + {x}^{2} } \\ \\ = \frac{ - 4}{5} \times \frac{1}{2} {tan}^{ - 1} \frac{x}{2} + \frac{9}{5} \times \frac{1}{3} tan ^{ - 1} \frac{x}{3} + c \\ \\ = \frac{ - 2}{5} {tan}^{ - 1} \frac{x}{2} + \frac{3}{2} {tan}^{ - 1} \frac{x}{3} + c$

Hope it helps uh