Let x2 = y

x2

(x2+4)(x2+9)

=

y

(y+)(y+9)

=

A

y+4

+

B

y+9

y = A (y + 9) + B (y + 4)
Comparing both sides,
A + B = 1 and 9A + 4B = 0
Solving, we get A =

−4

5

and B =

9

5

∴ I = ∫ |

−4

5(x2+4)

|+

9

5(x2+9)

dx
= −

4

5

×

1

2

tan−1(

x

2

) +

9

5

×

1

3

tan−1(

x

3

) + X
= -

2

5

tan−1

x

2

+

3

5

tan−1

x

3

+ C