Question

x +x+1x+1)2 (x+2)22.

Answer 1

The given function is x 2 +x+1 ( x+1 ) 2 ( x+2 ) .

Using method of partial fractions,

x 2 +x+1 ( x+1 ) 2 ( x+2 ) = A ( x+1 ) + B ( x+1 ) 2 + C ( x+2 ) x 2 +x+1=A( x+1 )( x+2 )+B( x+2 )+C ( x+1 ) 2 =A( x 2 +3x+2 )+B( x+2 )+C( x 2 +2x+1 ) =( A+C ) x 2 +( 3A+B+2C )x+( 2A+2B+C )

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

A+C=1(1)

3A+B+2C=1(2)

2A+2B+C=1(3)

On solving equations (1), (2) and (3), we get

A=−2 B=1 C=3

Now, the function becomes,

x 2 +x+1 ( x+1 ) 2 ( x+2 ) = −2 ( x+1 ) + 3 ( x+2 ) + 1 ( x+1 ) 2 (4)

Integrate equation (4),

∫ x 2 +x+1 ( x+1 ) 2 ( x+2 ) dx =−2 ∫ 1 x+1 dx +3 ∫ 2 x+2 dx + ∫ 1 ( x+1 ) 2 dx =−2 ∫ 1 x+1 dx +3 ∫ 2 x+2 dx + ∫ 1 ( x+1 ) 2 dx =−2log| x+1 |+3log| x+2 |− 1 x+1 +c

Thus, integration of given function is −2log| x+1 |+3log| x+2 |− 1 x+1 +c.