- #1

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## Homework Statement

How do i integrate:

[tex]\int\frac{1}{(x^2+1)^2}[/tex]

?

## Homework Equations

The answer is:

[tex]\frac{1}{4}(\frac{-2x}{x^2-1}-log(x-1)+log(x+1))[/tex]

## The Attempt at a Solution

I dont know which method to use.

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- Thread starter sara_87
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- #1

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How do i integrate:

[tex]\int\frac{1}{(x^2+1)^2}[/tex]

?

The answer is:

[tex]\frac{1}{4}(\frac{-2x}{x^2-1}-log(x-1)+log(x+1))[/tex]

I dont know which method to use.

- #2

Dick

Science Advisor

Homework Helper

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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

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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

Mark44

Mentor

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

- #5

Dick

Science Advisor

Homework Helper

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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

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

- #6

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how did you know what substitution to make??

- #7

Mark44

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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

Mark44

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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

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

- #10

Mark44

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

- #11

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thank you

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