How to solve for dx in the numerator by itself
$begingroup$
$$int frac{dx}{1+x^2} $$
I'm having a hard time knowing what to do when it's just dx on top
indefinite-integrals
edited Dec 10 '18 at 5:17
Patrick Stevens
28.6k52874
asked Dec 10 '18 at 5:14
A.BreeA.Bree
1
$endgroup$
add a comment |
$begingroup$
$$int frac{dx}{1+x^2} $$
I'm having a hard time knowing what to do when it's just dx on top
indefinite-integrals
edited Dec 10 '18 at 5:17
Patrick Stevens
28.6k52874
asked Dec 10 '18 at 5:14
A.BreeA.Bree
1
$endgroup$
add a comment |
$begingroup$
$$int frac{dx}{1+x^2} $$
I'm having a hard time knowing what to do when it's just dx on top
indefinite-integrals
edited Dec 10 '18 at 5:17
Patrick Stevens
28.6k52874
asked Dec 10 '18 at 5:14
A.BreeA.Bree
1
$endgroup$
$$int frac{dx}{1+x^2} $$
I'm having a hard time knowing what to do when it's just dx on top
indefinite-integrals
indefinite-integrals
edited Dec 10 '18 at 5:17
Patrick Stevens
28.6k52874
asked Dec 10 '18 at 5:14
A.BreeA.Bree
1
edited Dec 10 '18 at 5:17
Patrick Stevens
28.6k52874
asked Dec 10 '18 at 5:14
A.BreeA.Bree
1
edited Dec 10 '18 at 5:17
Patrick Stevens
28.6k52874
edited Dec 10 '18 at 5:17
Patrick Stevens
28.6k52874
edited Dec 10 '18 at 5:17
Patrick Stevens
28.6k52874
28.6k52874
asked Dec 10 '18 at 5:14
A.BreeA.Bree
1
asked Dec 10 '18 at 5:14
A.BreeA.Bree
1
asked Dec 10 '18 at 5:14
A.BreeA.Bree
1
1
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
If this was your confusion: the integral is precisely by definition equal to $$int frac{1}{1+x^2} mathrm{d}x$$
which is a very well-known integral.
answered Dec 10 '18 at 5:18
Patrick StevensPatrick Stevens
28.6k52874
$endgroup$
add a comment |
$begingroup$
$$int frac{1}{1+x^2} space dx = tan^{-1}(x) + C$$
answered Dec 10 '18 at 5:26
ArtArt
31217
$endgroup$
add a comment |
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2 Answers
2
active
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votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
If this was your confusion: the integral is precisely by definition equal to $$int frac{1}{1+x^2} mathrm{d}x$$
which is a very well-known integral.
answered Dec 10 '18 at 5:18
Patrick StevensPatrick Stevens
28.6k52874
$endgroup$
add a comment |
$begingroup$
If this was your confusion: the integral is precisely by definition equal to $$int frac{1}{1+x^2} mathrm{d}x$$
which is a very well-known integral.
answered Dec 10 '18 at 5:18
Patrick StevensPatrick Stevens
28.6k52874
$endgroup$
add a comment |
$begingroup$
If this was your confusion: the integral is precisely by definition equal to $$int frac{1}{1+x^2} mathrm{d}x$$
which is a very well-known integral.
answered Dec 10 '18 at 5:18
Patrick StevensPatrick Stevens
28.6k52874
$endgroup$
If this was your confusion: the integral is precisely by definition equal to $$int frac{1}{1+x^2} mathrm{d}x$$
which is a very well-known integral.
answered Dec 10 '18 at 5:18
Patrick StevensPatrick Stevens
28.6k52874
answered Dec 10 '18 at 5:18
Patrick StevensPatrick Stevens
28.6k52874
answered Dec 10 '18 at 5:18
Patrick StevensPatrick Stevens
28.6k52874
answered Dec 10 '18 at 5:18
Patrick StevensPatrick Stevens
28.6k52874
28.6k52874
add a comment |
add a comment |
$begingroup$
$$int frac{1}{1+x^2} space dx = tan^{-1}(x) + C$$
answered Dec 10 '18 at 5:26
ArtArt
31217
$endgroup$
add a comment |
$begingroup$
$$int frac{1}{1+x^2} space dx = tan^{-1}(x) + C$$
answered Dec 10 '18 at 5:26
ArtArt
31217
$endgroup$
add a comment |
$begingroup$
$$int frac{1}{1+x^2} space dx = tan^{-1}(x) + C$$
answered Dec 10 '18 at 5:26
ArtArt
31217
$endgroup$
$$int frac{1}{1+x^2} space dx = tan^{-1}(x) + C$$
answered Dec 10 '18 at 5:26
ArtArt
31217
answered Dec 10 '18 at 5:26
ArtArt
31217
answered Dec 10 '18 at 5:26
ArtArt
31217
answered Dec 10 '18 at 5:26
ArtArt
31217
31217
add a comment |
add a comment |
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