- #1
shan
- 57
- 0
I just want someone to check since I only learned this 'Integration using partial fractions' method today and I'm not sure I got it right...
First question:
[tex]\int \frac {dx}{x^2-a^2}[/tex] = [tex]\int \frac {dx}{(x+a)(x-a)}[/tex]
let [tex] \frac {1}{(x+a)(x-a)} \equiv \frac {A}{x+a} + \frac {B}{x-a}[/tex]
[tex]\equiv \frac {A(x-a)+B(x+a)}{(x+a)(x-a)}[/tex]
so [tex] 1 \equiv (A+B)x+a(-A+B)[/tex]
equate coefficients to find:
[tex]B=\frac{1}{2a}[/tex] and [tex]A=\frac{-1}{2a}[/tex]
substituting A and B back:
[tex]\int \frac {dx}{(x+a)(x-a)}[/tex] = -[tex]\int \frac {x+a}{2a} dx + \int \frac {x-a}{2a} dx[/tex]
I took the 1/2a out and integrated and got:
[tex]\frac {-1}{2a} \frac {x^2}{2} + ax + \frac {1}{2a} \frac {x^2}{2} - ax + c[/tex]
which simplifies to c which looks extremely strange to me...
Second question:
[tex]\int \frac {dx}{16x^4-1} = \int \frac {dx}{(4x^2+1)(2x+1)(2x-1)}[/tex]
let [tex]\frac {1}{16x^4-1} \equiv \frac {Ax+B}{4x^2+1} + \frac {C}{2x+1} + \frac {D}{2x-1}[/tex]
[tex]\equiv \frac {(Ax+B)(4x^2-1)+C(2x-1)(4x^2+1)+D(2x+1)(4x^2+1)}{16x^4-1}[/tex]
[tex]1 \equiv x^3(4A+8C+8D)+x^2(4B-4C+4D)+x(-A+2C+2D)+D-C-B[/tex]
equating coffecients to get:
4A+8C+8D=0
4B-4C+4D=0
-A+2C+2D=0
D-C-B=1
I then found:
A=0, B=-1/2, C=-1/4 and D=1/4
substituting back into the integral:
[tex]\int \frac {dx}{16x^4-1} = \frac {-1}{2} \int 4x^2+1 dx - \frac {1}{4} \int 2x+1 dx + \frac {1}{4} \int 2x+1[/tex]
= [tex]\frac{-4x^3}{6}+x-\frac{x^2}{4}+x+\frac{x^2}{4}-x+c[/tex]
= [tex]\frac {-2x^3}{3}+x+c[/tex]
Last question:
[tex]\int \frac {2}{(x^2+x+2)(x+1)}dx[/tex]
let [tex]\frac {2}{(x^2+x+2)(x+1)} \equiv \frac {Ax+B}{(x+\frac {1}{2})^2 + \frac {7}{4}} + \frac {C}{x+1}[/tex]
[tex]\equiv \frac {(Ax+B)(x+1)+C((x+\frac {1}{2})^2+\frac {7}{4})}{(x^2+x+2)(x+1)}[/tex]
[tex]2\equiv x^2(A+C)+x(A+B+C)+B+2C[/tex]
equating coefficients:
A+C=0
A+B+C=0
B+2C=2
so: A=-1, B=0, C=1
putting it back into the integral:
[tex]\int \frac {2}{(x^2+x+2)(x+1)}dx = -\int \frac {x}{(x+\frac {1}{2})^2+\frac {7}{4}} dx + \int \frac {1}{x+1} dx[/tex]
I integrated the first part using the 'Integration by parts' method and then simplified to get:
[tex]\frac {-x}{\sqrt {\frac{7}{4}}} \arctan (\frac {x+\frac{1}{2}}{\sqrt {7}}) - \tan (\frac {2x+1}{\sqrt{7}}) + ln(x+1) + c[/tex]
Thanks in advance :)
First question:
[tex]\int \frac {dx}{x^2-a^2}[/tex] = [tex]\int \frac {dx}{(x+a)(x-a)}[/tex]
let [tex] \frac {1}{(x+a)(x-a)} \equiv \frac {A}{x+a} + \frac {B}{x-a}[/tex]
[tex]\equiv \frac {A(x-a)+B(x+a)}{(x+a)(x-a)}[/tex]
so [tex] 1 \equiv (A+B)x+a(-A+B)[/tex]
equate coefficients to find:
[tex]B=\frac{1}{2a}[/tex] and [tex]A=\frac{-1}{2a}[/tex]
substituting A and B back:
[tex]\int \frac {dx}{(x+a)(x-a)}[/tex] = -[tex]\int \frac {x+a}{2a} dx + \int \frac {x-a}{2a} dx[/tex]
I took the 1/2a out and integrated and got:
[tex]\frac {-1}{2a} \frac {x^2}{2} + ax + \frac {1}{2a} \frac {x^2}{2} - ax + c[/tex]
which simplifies to c which looks extremely strange to me...
Second question:
[tex]\int \frac {dx}{16x^4-1} = \int \frac {dx}{(4x^2+1)(2x+1)(2x-1)}[/tex]
let [tex]\frac {1}{16x^4-1} \equiv \frac {Ax+B}{4x^2+1} + \frac {C}{2x+1} + \frac {D}{2x-1}[/tex]
[tex]\equiv \frac {(Ax+B)(4x^2-1)+C(2x-1)(4x^2+1)+D(2x+1)(4x^2+1)}{16x^4-1}[/tex]
[tex]1 \equiv x^3(4A+8C+8D)+x^2(4B-4C+4D)+x(-A+2C+2D)+D-C-B[/tex]
equating coffecients to get:
4A+8C+8D=0
4B-4C+4D=0
-A+2C+2D=0
D-C-B=1
I then found:
A=0, B=-1/2, C=-1/4 and D=1/4
substituting back into the integral:
[tex]\int \frac {dx}{16x^4-1} = \frac {-1}{2} \int 4x^2+1 dx - \frac {1}{4} \int 2x+1 dx + \frac {1}{4} \int 2x+1[/tex]
= [tex]\frac{-4x^3}{6}+x-\frac{x^2}{4}+x+\frac{x^2}{4}-x+c[/tex]
= [tex]\frac {-2x^3}{3}+x+c[/tex]
Last question:
[tex]\int \frac {2}{(x^2+x+2)(x+1)}dx[/tex]
let [tex]\frac {2}{(x^2+x+2)(x+1)} \equiv \frac {Ax+B}{(x+\frac {1}{2})^2 + \frac {7}{4}} + \frac {C}{x+1}[/tex]
[tex]\equiv \frac {(Ax+B)(x+1)+C((x+\frac {1}{2})^2+\frac {7}{4})}{(x^2+x+2)(x+1)}[/tex]
[tex]2\equiv x^2(A+C)+x(A+B+C)+B+2C[/tex]
equating coefficients:
A+C=0
A+B+C=0
B+2C=2
so: A=-1, B=0, C=1
putting it back into the integral:
[tex]\int \frac {2}{(x^2+x+2)(x+1)}dx = -\int \frac {x}{(x+\frac {1}{2})^2+\frac {7}{4}} dx + \int \frac {1}{x+1} dx[/tex]
I integrated the first part using the 'Integration by parts' method and then simplified to get:
[tex]\frac {-x}{\sqrt {\frac{7}{4}}} \arctan (\frac {x+\frac{1}{2}}{\sqrt {7}}) - \tan (\frac {2x+1}{\sqrt{7}}) + ln(x+1) + c[/tex]
Thanks in advance :)