 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Interesting Integral Illation (Posted on 2023-04-06) ```Evaluate this definite integral:

n
∫ x2*(1+x4)-1 dx
1/n

where n is a real number greater than 1.
```

 See The Solution Submitted by K Sengupta Rating: 5.0000 (1 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Some analytic progress | Comment 2 of 3 | Integ {1/n, n} x^2/(x^4+1) dx

Let's start with the substitution y=1/x.  then dy=-1/(x^2) n->1/n and 1/n->n. Then after some simplification we have
Integ {1/n, n} 1/(y^4+1) dy
Since this is a definite integral, y is just a dummy variable and can be changed back to x.

This result is very useful when we combine the two integrals by taking their average to make
1/2 * Integ {1/n, n} (x^2+1)/(x^4+1) dx

(x^2+1)/(x^4+1) has a rather simple partial fraction decomposition of
(1/2)/(x^2+sqrt(2)x+1) + (1/2)/(x^2-sqrt(2)x+1)

Then I'll complete the square and substitute this into the integral
1/4 * Integ {1/n, n} 1/[(x+1/sqrt(2))^2+(1/sqrt(2))^2] + 1/[(x-1/sqrt(2))^2+(1/sqrt(2))^2] dx

Now both terms of the integral can be integrated using arctan.  Doing so and simplifying a bit yields
1/(2*sqrt(2)) * (arctan[sqrt(2)*x+1] + arctan[sqrt(2)*x-1]) |{1/n, n}

At this point I want to apply the arctan addition formula to simplify stuff, but it doesn't quite work cleanly when the arguments are outside of the interval (-1,1)

 Posted by Brian Smith on 2023-04-07 11:40:06 Please log in:

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