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Anti-derivative help?
(2+x^2)/(1+x^2)
i know the denominator anti-derivative is tan^-1 x
how do i get to this?
2 Answers
- BuriLv 71 decade agoFavorite Answer
Notice that it can be written like this:
[1 + (1+x^2)]/1+x^2 = 1/(1+x^2) + (1+x^2)/(1+x^2) = 1/(1+x^2) +1
And so anti-derivative is:
arctan(x) + x +C
Hope this helps!
- PuggyLv 71 decade ago
Integral ( [(2 + x^2)/(1 + x^2)] dx )
Since the degree of the numerator is greater than or equal to the degree of the denominator, you can use long division to decompose into two fractions. However, there is a simpler way.
First, express 2 as 1 + 1.
Integral ( [ 1 + 1 + x^2 ] / (1 + x^2) ] dx )
Next, split into two fractions with a common denominator.
Integral ( [ 1/(1 + x^2) + (1 + x^2)/(1 + x^2) ] dx )
Note how we can simplify the second fraction to 1.
Integral ( [ 1/(1 + x^2) + 1 ] dx )
And we can integrate each individually. 1/(1 + x^2) is a known derivative; it is the derivative of arctan(x). Also, 1 is easy to integrate.
arctan(x) + x + C
