Need help checking integration answers

[shan]

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:
\int \frac {dx}{x^2-a^2} = \int \frac {dx}{(x+a)(x-a)}

let \frac {1}{(x+a)(x-a)} \equiv \frac {A}{x+a} + \frac {B}{x-a}

\equiv \frac {A(x-a)+B(x+a)}{(x+a)(x-a)}

so 1 \equiv (A+B)x+a(-A+B)

equate coefficients to find:
B=\frac{1}{2a} and A=\frac{-1}{2a}

substituting A and B back:
\int \frac {dx}{(x+a)(x-a)} = -\int \frac {x+a}{2a} dx + \int \frac {x-a}{2a} dx

I took the 1/2a out and integrated and got:
\frac {-1}{2a} \frac {x^2}{2} + ax + \frac {1}{2a} \frac {x^2}{2} - ax + c

which simplifies to c which looks extremely strange to me...


Second question:
\int \frac {dx}{16x^4-1} = \int \frac {dx}{(4x^2+1)(2x+1)(2x-1)}

let \frac {1}{16x^4-1} \equiv \frac {Ax+B}{4x^2+1} + \frac {C}{2x+1} + \frac {D}{2x-1}

\equiv \frac {(Ax+B)(4x^2-1)+C(2x-1)(4x^2+1)+D(2x+1)(4x^2+1)}{16x^4-1}

1 \equiv x^3(4A+8C+8D)+x^2(4B-4C+4D)+x(-A+2C+2D)+D-C-B

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:
\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
= \frac{-4x^3}{6}+x-\frac{x^2}{4}+x+\frac{x^2}{4}-x+c
= \frac {-2x^3}{3}+x+c


Last question:
\int \frac {2}{(x^2+x+2)(x+1)}dx

let \frac {2}{(x^2+x+2)(x+1)} \equiv \frac {Ax+B}{(x+\frac {1}{2})^2 + \frac {7}{4}} + \frac {C}{x+1}

\equiv \frac {(Ax+B)(x+1)+C((x+\frac {1}{2})^2+\frac {7}{4})}{(x^2+x+2)(x+1)}

2\equiv x^2(A+C)+x(A+B+C)+B+2C

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:
\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

I integrated the first part using the 'Integration by parts' method and then simplified to get:

\frac {-x}{\sqrt {\frac{7}{4}}} \arctan (\frac {x+\frac{1}{2}}{\sqrt {7}}) - \tan (\frac {2x+1}{\sqrt{7}}) + ln(x+1) + c

Thanks in advance :)

 

[Data]

For #1 and #2, I want you to look very hard at the part where you subbed A and B back in and see if you see anything wrong (are you sure all those functions of x should be in the numerator?).

I'll look at 3 in a minute~

 

[Data]

3 looks fine from the work you have down, but your answer is wrong. Check over your integration by parts~

 

[shan]

Oops...

is #1
\frac {-1}{2a} ln(2ax+2a^2) + \frac {1}{2a} ln(2ax-2a^2) + c<br /> ?<br /> <br /> for #2<br /> \frac {-1}{2} \int \frac {1}{4x^2+1} dx - \frac {1}{8}ln(2x+1) + \frac {1}{8}ln(2x+1)<br /> I don&#039;t know how to integrate \int \frac {1}{4x^2+1} dx ?<br /> <br /> #3...I got to:<br /> \frac {-x}{2(x^2+x+2)} - \int \frac {1}{2x^4+8x^2+8x+4}<br /> but I don&#039;t know what to do after that...

 

[shan]

sorry, sorry, I mucked up the reply...

it's supposed to be...



Oops...

is #1
\frac {-1}{2a} ln(2ax+2a^2) + \frac {1}{2a} ln(2ax-2a^2) + c
?

for #2
\frac {-1}{2} \int \frac {1}{4x^2+1} dx - \frac {1}{8}ln(2x+1) + \frac {1}{8}ln(2x+1)
I don't know how to integrate \int \frac {1}{4x^2+1} dx though?

#3...I got to:
\frac {-x}{2(x^2+x+2)} - \int \frac {1}{2x^4+8x^2+8x+4}
but I don't know what to do after that...

 

[dextercioby]

\int \frac{dx}{4x^{2}+1}=\frac{1}{2}\int \frac{d(2x)}{(2x)^{2}+1}=\frac{1}{2}\arctan 2x +C

Okay...?

 

[shan]

so #2 becomes
\frac {-1}{4} \arctan(2x) - \frac {1}{8}ln(2x+1) + \frac {1}{8}ln(2x+1) + c ?

 

[Data]

#1 looks right. You can simplify it to

\frac{1}{2a} \ln \left(\frac{x-a}{x+a} \right) + C

too. Do you see how?

#2 is perfect.

For 3, note that

\int \frac{2}{(x^2+x+2)(x+1)} \ dx = -\int \frac{x}{\left(x+\frac{1}{2}\right)^2 + \frac{7}{4}} \ dx + \int \frac{dx}{x+1}

= -\int \frac{x + \frac{1}{4}}{\left(x + \frac{1}{2}\right)^2 + \frac{7}{4}} \ dx + \int \frac{\frac{1}{4}}{\left(x + \frac{1}{2}\right)^2 + \frac{7}{4}} \ dx + \ln (x+1)

= -\int \frac{x + \frac{1}{4}}{\left(x + \frac{1}{2}\right)^2 + \frac{7}{4}} \ dx + \frac{1}{2\sqrt{7}}\arctan \left( \frac{2x+1}{\sqrt{7}} \right) + \ln (x+1)

and see if the substitution u = (x+ 1/2)^2 will help to evaluate the first integral.