# Need help checking integration answers

1. Mar 20, 2005

### shan

I just want someone to check since I only learnt 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:

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

2. Mar 21, 2005

### 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~

3. Mar 21, 2005

### Data

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

4. Mar 21, 2005

### shan

Oops....

is #1
$$\frac {-1}{2a} ln(2ax+2a^2) + \frac {1}{2a} ln(2ax-2a^2) + c ? for #2 [tex]\frac {-1}{2} \int \frac {1}{4x^2+1} dx - \frac {1}{8}ln(2x+1) + \frac {1}{8}ln(2x+1)$$
I don't know how to integrate $$\int \frac {1}{4x^2+1} dx$$ ?

#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...

5. Mar 21, 2005

### 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...

6. Mar 21, 2005

### dextercioby

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

Okay...?

7. Mar 21, 2005

### shan

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

8. Mar 21, 2005

### 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.

Last edited: Mar 21, 2005