# Integration Problem

1. Nov 16, 2009

### sara_87

1. The problem statement, all variables and given/known data

How do i integrate:

$$\int\frac{1}{(x^2+1)^2}$$

?

2. Relevant equations

$$\frac{1}{4}(\frac{-2x}{x^2-1}-log(x-1)+log(x+1))$$

3. The attempt at a solution

I dont know which method to use.

2. Nov 16, 2009

### Dick

Re: Integration

The answer you posted is correct if the integrand is 1/(1-x^2)^2. Is that what you meant? If so, try partial fractions.

3. Nov 16, 2009

### sara_87

Re: Integration

Im sorry, the question is correct but the answer is:

1/2[x/(x^2+1)+arctan(x)]

here partial fractions wont work because i will still have a term with (x^2+1)^2 in the denominator

4. Nov 16, 2009

### Staff: Mentor

Re: Integration

Have you tried a trig substitution? tan (theta) = x would be a good one to start with.

5. Nov 16, 2009

### Dick

Re: Integration

Ok, then it's a trig substitution. Like x=tan(t).

6. Nov 16, 2009

### sara_87

Re: Integration

how did you know what substitution to make??

7. Nov 16, 2009

### Staff: Mentor

Re: Integration

Draw a right triangle and label the sides and hypotenuse according to the expression in your integral. Since you have x^2 + 1, that corresponds to the hypotenuse. Label the side opposite your acute angle as x, label the base as 1, and label the hypotenuse as sqrt(x^2 + 1). So tan(theta) = x/1, and sec^2(theta)d(theta) = dx. Use these two equations to completely convert you integral from expressions in x and dx to ones in theta and d(theta).

8. Nov 16, 2009

### Staff: Mentor

Re: Integration

If you're asking how did we know to make a trig substitution, they are good bets when you have factors involving the sum or difference of squares, such as sqrt(x^2 + a^2), sqrt(a^2 - x^2), or sqrt(x^2 - a^2).

9. Nov 16, 2009

### sara_87

Re: Integration

what if we have (5+x^2)^2 in the denominator?

10. Nov 16, 2009

### Staff: Mentor

Re: Integration

Label the opposite side as x and the base as sqrt(x), so tan(theta) = x/sqrt(5)

11. Nov 16, 2009

### sara_87

Re: Integration

thank you