Question

How do you find `\int \frac { 2x + 5} { x ^ { 3} + 6x ^ { 2} + 9x } d y`?

Answer 1

`int (2x+5)/(x^3+6x^2+9)dx = 5/9 ln abs x -5/9 ln abs(x+3) -1/3 1 /(x+3)+C`

Explanation:

Factorize the denominator:

`x^3+6x^2+9 = x(x+3)^2`

then decompose the rational function in partial fractions:

`(2x+5)/(x^3+6x^2+9) = A/x +B/(x+3)+ C/(x+3)^2`

`(2x+5)/(x^3+6x^2+9) = (A(x+3)^2 +Bx(x+3)+ Cx)/(x^3+6x^2+9)`

Equating the numerators:

`2x +5 = Ax^2 +6Ax +9A + Bx^2 +3Bx +Cx`

`2x +5 = (A+B)x^2 + (6A+3B+C) x + 9A`

Finally equation the coefficients of the same degree in `x`:

`{(A+B = 0),(6A+3B+C= 2), ( 9A = 5):}`

`{(A=5/9),(B=-5/9),(C = 1/3):}`

Then:

`int (2x+5)/(x^3+6x^2+9)dx = 5/9 int dx/x -5/9 int dx/(x+3) +1/3 int dx/(x+3)^2`

`int (2x+5)/(x^3+6x^2+9)dx = 5/9 ln abs x -5/9 ln abs(x+3) -1/3 1 /(x+3)+C`