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How do you integrate #int(4x)/(x^4 +1) # using partial fractions?

Answer 1

Emilia Aguilar

#2arctan(x^2) +c#

I believe that you can't use partial fractions because #Delta<0# and you can't decompose the denominator

Use this immediate integral: #int(f'(x))/(1+[f(x)]^2) = arctan[f(x)] + c#

Write your integral in this way: #2int(2x)/(1+[(x^2)]^2)#

So the result is: #2arctan(x^2) +c#

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Answer 2

Charles Anctil

To integrate ( \int \frac{4x}{x^4 + 1} ) using partial fractions, first factor the denominator:

( x^4 + 1 = (x^2 + 1)^2 - 2x^2 = (x^2 - \sqrt{2}x + 1)(x^2 + \sqrt{2}x + 1) )

Now, we can express ( \frac{4x}{x^4 + 1} ) as partial fractions:

( \frac{4x}{x^4 + 1} = \frac{Ax + B}{x^2 - \sqrt{2}x + 1} + \frac{Cx + D}{x^2 + \sqrt{2}x + 1} )

By comparing numerators, we get:

( 4x = (Ax + B)(x^2 + \sqrt{2}x + 1) + (Cx + D)(x^2 - \sqrt{2}x + 1) )

Solve for ( A ), ( B ), ( C ), and ( D ) using algebraic manipulation.

Once you find the values of ( A ), ( B ), ( C ), and ( D ), you can integrate each term separately. The integrals will likely involve natural logarithms and arctangent functions.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.