Question

`int( (y^2 + 2 y - 2)dy)/(y^3 + 4 y^2 - 7 y - 10)=` ?

Answer 1

See below.

Explanation:

`y^3+4y^2-7y-10 = (y+5)(y+1)(y-2)` then

`(y^2 + 2 y - 2)/(y^3 + 4 y^2 - 7 y - 10) = a/(y+5)+b/(y+1)+c/(y+2)` and finally

`int( (y^2 + 2 y - 2)dy)/(y^3 + 4 y^2 - 7 y - 10)=a int dy/(y+5) + b int dy/(y+1)+c int dy/(y+2) + C`

where

`a = lim_(y->-5)((y^2 + 2 y - 2)/(y^3 + 4 y^2 - 7 y - 10))(y+5) = 13/28`
`b = lim_(y->-1)((y^2 + 2 y - 2)/(y^3 + 4 y^2 - 7 y - 10))(y+1) = 1/4`
`c = lim_(y->2)((y^2 + 2 y - 2)/(y^3 + 4 y^2 - 7 y - 10))(y-2)=2/7`

then

`int( (y^2 + 2 y - 2)dy)/(y^3 + 4 y^2 - 7 y - 10)=13/28 log_eabs(y+5)+1/4 log_e abs(y+1)+2/7 log_e abs(y-2) + C`

Answer 2

`int(y^2+2y-2)/(y^3-4y^2-7y-10)dy = 13/28ln|y+5|+1/4ln|y+1|+2/7ln|y-2|+C`

Explanation:

The factors of `(y^3-4y^2-7y-10) = (y+5)(y+1)(y-2)`

Decompose the integrand into partial fractions:

`(y^2+2y-2)/(y^3-4y^2-7y-10)= A/(y+5)+B/(y+1)+C/(y-2)`

Multiply both sides by `(y+5)(y+1)(y-2)`:

`y^2+2y-2 = A(y+1)(y-2)+B(y+5)(y-2)+C(y+5)(y+1)`

Make the B and C terms disappear by letting `y = -5`:

`(-5)^2+2(-5)-2 = A((-5)+1)((-5)-2)`

`13=A(-4)(-7)`

A = 13/28

Make the A and C terms disappear by letting `y =1`:

`(-1)^2+2(-1)-2 = B((-1)+5)((-1)-2)`

`-3 = B(4)(-3)`

`B = 1/4`

Make the A and B terms disappear by letting `y = 2`

`2^2+2(2)-2 = C((2)+5)((2)+1)`

`6 = C(7)(3)`

`C = 2/7`

Therefore, the integral becomes 3 integrals with these coefficients:

`int(y^2+2y-2)/(y^3-4y^2-7y-10)dy = 13/28int1/(y+5)dy+1/4int1/(y+1)dy+2/7int1/(y-2)dy`

These all integrate into natural logarithms:

`int(y^2+2y-2)/(y^3-4y^2-7y-10)dy = 13/28ln|y+5|+1/4ln|y+1|+2/7ln|y-2|+C`