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How do you use Integration by Substitution to find #intx/(x^4+1)dx#?

Answer 1

Let #u=x^2#. By taking the derivative, #{du}/{dx}=2x# by taking the reciprocal, #Rightarrow{dx}/{du}=1/{2x}# by multiplying by #du#, #Rightarrow dx={du}/{2x}#

By rewriting the integral in terms of #u#, #intx/{x^4+1}dx=intx/{u^2+1}cdot{du}/{2x}# by cancelling out #x#'s, #=1/2 int1/{1+u^2}du=1/2tan^{-1}u+C# by putting #u=x^2# back in, #=1/2tan^{-1}(x^2)+C#

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Answer 2

Emma Aldridge

User is interested in mathematics, specifically calculus.To find the integral of ( \frac{x}{x^4 + 1} ) with respect to ( x ) using integration by substitution, we can let ( u = x^2 ). Then, ( du = 2x dx ). Rearranging this gives us ( \frac{1}{2} du = x dx ). Substituting these into the integral, we have:

[ \int \frac{x}{x^4 + 1} dx = \frac{1}{2} \int \frac{1}{u^2 + 1} du ]

Now, we recognize that ( \frac{1}{u^2 + 1} ) is the derivative of ( \arctan(u) ). So, the integral becomes:

[ \frac{1}{2} \arctan(u) + C ]

Substituting back ( u = x^2 ), we get the final result:

[ \frac{1}{2} \arctan(x^2) + C ]

where ( C ) is the constant of integration.

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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.