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적분 공식
Basic Forms
1 | ∫u dv=uv−∫v du | 2 | ∫un du=1n+1un+1+C , n≠−1 |
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3 | ∫1u du=ln |u|+C | 4 | ∫eu du=en+C |
5 | ∫au du=1ln aau+C | 6 | ∫sin u du=−cos u+C |
7 | ∫cos u du=sin u+C | 8 | ∫sec2u du=tan u+C |
9 | ∫csc2u du=−cot u+C | 10 | ∫sec u tan u du=csc u+C |
11 | ∫csc u cot u du=−csc u+C | 12 | ∫tan u du=ln |sec u|+C |
13 | ∫cot u du=ln |sin u|+C | 14 | ∫sec u du=ln |sec u+tan u|+C |
15 | ∫csc u du=ln |csc u−cot u|+C | 16 | ∫1√a2−u2 du=sin−1ua+C |
17 | ∫1a2+u2 du=1atan−1ua+C | 18 | ∫1u√u2−a2 du=1asec−1ua+C |
19 | ∫1a2−u2 du=12aln |u+au−a|+C | 20 | ∫1u2−a2 du=12aln |u−au+a|+C |
Forms Involving 1
√a2+u2 , a>0
1 | ∫√a2+u2 du=u2√a2+u2+a22ln(u+√a2+u2)+C |
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2 | ∫u2√a2+u2 du=u8(a2+2u2)√a2+u2−a48ln(u+√a2+u2)+C |
3 | ∫√a2+u2u du=√a2+u2−a ln |a+√a2+u2u|+C |
4 | ∫√a2+u2u2 du=−√a2+u2u+ln(u+√a2+u2)+C |
5 | ∫1√a2+u2 du=ln(u+√a2+u2)+C |
6 | ∫u2√a2+u2 du=u2√a2+u2−a22ln(u+√a2+u2)+C |
8 | ∫1u√a2+u2 du=−1aln|√a2+u2+au|+C |
9 | ∫1u2√a2+u2 du=−√a2+u2a2u+C |
10 | ∫1(a2+u2)3/2 du=ua2√a2+u2+C |
Forms Involving 2
√a2−u2 , a>0
1 | ∫√a2−u2 du=u2√a2−u2+a22sin−1ua+C |
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2 | ∫u2√a2−u2 du=u8(2u2−a2)√a2−u2+a48sin−1ua+C |
3 | ∫√a2−u2u du=√a2−u2−a ln|a+√a2−u2u|+C |
4 | ∫√a2−u2u2 du=−1u√a2−u2−sin−1ua+C |
5 | ∫u2√a2−u2 du=−u2√a2−u2+a22sin−1ua+C |
6 | ∫1u√a2−u2 du=−1aln|a+√a2−u2u|+C |
7 | ∫1u2√a2−u2 du=−1a2u√a2−u2+C |
8 | ∫(a2−u2)3/2 du=−u8(2u2−5a2)√a2−u2+3a48sin−1ua+C |
9 | ∫1(a2−u2)3/2=ua2√a2−u2+C |
Forms Involving 3
√u2−a2 , a>0
1 | ∫√u2−a2 du=u2√u2−a2−a22ln |u+√u2−a2|+C |
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2 | ∫u2√u2−a2 du=u8(2u2−a2)√u2−a2−a48ln |u+√u2−a2|+C |
3 | ∫√u2−a2u du=√u2−a2−a cos−1a|u|+C |
4 | ∫√u2−a2u2 du=−√u2−a2u+ln |u+√u2−a2|+C |
5 | ∫1√u2−a2 du=ln |u+√u2−a2|+C |
6 | ∫u2√u2−a2 du=u2√u2−a2+a22ln |u+√u2−a2|+C |
7 | ∫1u2√u2−a2 du=√u2−a2a2u+C |
8 | ∫1(u2−a2)3/2 du=−ua2√u2−a2+C |
Forms Involving 4
a+bu
1 | ∫ua+bu du=1b2(a+bu−aln|a+bu|)+C | |||||||||||
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2 | ∫u2a+bu du=12b3[(a+bu)2−4a(a+bu)+2a2ln|a+bu|]+C | |||||||||||
3 | ∫1u(a+bu) du=1aln|ua+bu|+C | |||||||||||
4 | ∫1u2(a+bu) du=−1au+ba2ln|ua+bu|+C | |||||||||||
5 | ∫u(a+bu)2 du=ab2(a+bu)+1b2ln|a+bu|+C | |||||||||||
6 | ∫1u(a+bu)2 du=1a(a+bu)−1a2ln|a+buu|+C | |||||||||||
7 | ∫u2(a+bu)2 du=1b3(a+bu−a2a+bu−2aln|a+bu|)+C | |||||||||||
8 | ∫u√a+bu du=115b2(3bu−2a)(a+bu)3/2+C | |||||||||||
9 | ∫u√a+bu du=23b2(bu−2a)√a+bu+C | |||||||||||
10 | ∫u2√a+bu du=115b2(8a2+3b2u2−4abu)√a+bu+C | |||||||||||
11 | \begin{displaymath}\begin{split} \int \frac{1}{u \sqrt{a+bu}} \ du &= \frac{1}{\sqrt{a}} \ln \left| \frac{\sqrt{a+bu} - \sqrt{a}}{\sqrt{a+bu} + \sqrt{a}} \right| + C \ \ , \ if \ a > 0 \\ &= \frac{2}{\sqrt{-a}} \tan^{-1} \sqrt{\frac{a+bu}{-a}} + C \ \ \ \ , \ if \ a < 0 \end{split}\end{displaymath} | ^ 12 |∫√a+buu du=2√a+bu+a ∫1u√a+bu du | ^ 13 |$$ \int \frac{\sqrt{a+bu}}{u | {2}} \ du = - \frac{\sqrt{a+bu}}{u} + \frac{b}{2} \ \int \frac{1}{u \sqrt{a+bu}} \ du |14| \int u | {n} \sqrt{a+bu} \ du = \frac{2}{b(2n+3)} \left[ u | {n} (a+bu) | {3/2} -na \ \int u | {n-1} \sqrt{a+bu} \ du \right] |15| \int \frac{u | {n}}{\sqrt{a+bu}} \ du = \frac{2u | {n} \sqrt{a+bu}}{b(2n+1)} - \frac{2na}{b(2n+1)} \ \int \frac{u | {n-1}}{\sqrt{a+bu}} \ du |16| \int \frac{1}{u | {n} \sqrt{a+bu}} \ du = - \frac{\sqrt{a+bu}}{a(n-1) u | {n-1}} - \frac{b(2n-3)}{2a(n-1)} \ \int \frac{1}{u | {n-1} \sqrt{a+bu}} \ du $$ |