We are asked to find `int (x-15)/((x^2+2x+5)(x^2+6x+10))dx`
We will use partial fractions to rewrite the integrand:
`(x-15)/((x^2+2x+5)(x^2+6x+10))=(Ax+B)/(x^2+2x+5)+(Cx+D)/(x^2+6x+10)`
Now multiply both sides of the equation by the LCD to get:
`x-15=(Ax+B)(x^2+6x+10)+(Cx+D)(x^2+2x+5)`
Distributing and collecting like terms we get:
`x-15=(A+C)x^3+(6A+B+2C+D)x^2+(10A+6B+5C+2D)x+(10B+5D)`
Equating the coefficients we get A+C=0,6A+B+2C+D=0
10A+6B+5C+2D=1 and 10B+5D=-15
Solving the system yields A=B=1,C=-1,D=-5. Thus
`int (x+1)/(x^2+2x+5)dx-int (x+5)/(x^2+6x+10)dx`
The first integral can be done by letting `u=x^2+2x+5` and `du=2x+2 dx ` so
` int (x+1)/(x^2+2x+5)dx=1/2int (du)/u = 1/2...
We are asked to find `int (x-15)/((x^2+2x+5)(x^2+6x+10))dx`
We will use partial fractions to rewrite the integrand:
`(x-15)/((x^2+2x+5)(x^2+6x+10))=(Ax+B)/(x^2+2x+5)+(Cx+D)/(x^2+6x+10)`
Now multiply both sides of the equation by the LCD to get:
`x-15=(Ax+B)(x^2+6x+10)+(Cx+D)(x^2+2x+5)`
Distributing and collecting like terms we get:
`x-15=(A+C)x^3+(6A+B+2C+D)x^2+(10A+6B+5C+2D)x+(10B+5D)`
Equating the coefficients we get A+C=0,6A+B+2C+D=0
10A+6B+5C+2D=1 and 10B+5D=-15
Solving the system yields A=B=1,C=-1,D=-5. Thus
`int (x+1)/(x^2+2x+5)dx-int (x+5)/(x^2+6x+10)dx`
The first integral can be done by letting `u=x^2+2x+5` and `du=2x+2 dx ` so
` int (x+1)/(x^2+2x+5)dx=1/2int (du)/u = 1/2 ln|x^2+2x+5| `
Rewrite the second integral as:
`int (x+5)/(x^2+6x+10)dx = int (2x+6)/(x^2+6x+10)dx - 4int (dx)/(x^2+6x+10)`
Then we get:
`1/2 ln|x^2+6x+10|`
and
`-4 int (dx)/((x+3)^2+1)=-4tan^(-1)(x+3)`
Thus:
`int (x-15)/((x^2+2x+5)(x^2+6x+10))dx='`
`1/2(ln|x^2+2x+5|-ln|x^2+6x+10|)-4tan^(-1)(x+3)+C
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