Question

5x(x +1) (r2 +96.

Answer 1

The given function f( x ) is 5x ( x+1 )( x 2 +9 ) .

Using method of partial fractions,

5x ( x+1 )( x 2 +9 ) = A x+1 + Bx+C x 2 +9 5x=A( x 2 +9 )+( Bx+C )( x+1 ) =A x 2 +9A+B x 2 +Bx+Cx+C

Equating the coefficients of x, x 2 and constant terms,

A+B=0 A=−B (1)

B+C=5(2)

9A+C=0(3)

From, equation (1) and (3),

−9B+C=0 C=9B (4)

From equation (2) and (4)

9B+B=5 10B=5 B= 5 10 = 1 2 (5)

From equation (1) and (5)

A=−B =− 1 2

From equation (1) and (5)

C=9B = 9 2

Now, the function is,

5x ( x+1 )( x 2 +9 ) = −1 2 x+1 + 1 2 x+ 9 2 x 2 +9 .(6)

Integrate equation (6).

∫ 5x ( x+1 )( x 2 +9 ) dx =− −1 2 ∫ dx x+1 + 1 2 ∫ xdx x 2 +9 + 9 2 ∫ dx x 2 +9 = −1 2 log| x+1 |+ 1 4 ∫ 2xdx x 2 +9 + 9 2 ⋅ 1 3 tan −1 x 3 +c = −1 2 log| x+1 |+ 1 4 log| x 2 +9 |+ 3 2 tan −1 x 3 +c

Thus, integration of f( x ) is −1 2 log| x+1 |+ 1 4 log| x 2 +9 |+ 3 2 tan −1 x 3 +c.