Pagrindiniai ir dažniausiai pasitaikantys integralai :
∫ 0 d x = C {\displaystyle \int 0\;{\mathsf {d}}x=C} ∫ a d x = a x + C {\displaystyle \int a\;{\mathsf {d}}x=ax+C} ∫ x n d x = x n + 1 n + 1 + C {\displaystyle \int x^{n}\;{\mathsf {d}}x={\frac {x^{n+1}}{n+1}}+C} ∫ n d x x = n ln | x | + C {\displaystyle \int {\frac {{\mathsf {nd}}x}{x}}=n\ln \left|x\right|+C} ∫ e x d x = e x + C {\displaystyle \int {\mathsf {e}}^{x}\;{\mathsf {d}}x={\mathsf {e}}^{x}+C} ∫ a x d x = a x ln a + C {\displaystyle \int a^{x}\;{\mathsf {d}}x={\frac {a^{x}}{\ln a}}+C} ∫ d x x 2 + a 2 = 1 a arctan x a + C , a ≠ 0 {\displaystyle \int {\frac {{\mathsf {d}}x}{x^{2}+a^{2}}}={\frac {1}{a}}\arctan {\frac {x}{a}}+C,a\not =0} ∫ 1 a 2 − x 2 d x = arcsin x a + C , a > 0 {\displaystyle \int {\frac {1}{\sqrt {a^{2}-x^{2}}}}\;{\mathsf {d}}x=\arcsin {\frac {x}{a}}+C,\;a>0} ∫ d x x 2 − a 2 = 1 2 a ln | x − a x + a | + C , a ≠ 0 {\displaystyle \int {\frac {{\mathsf {d}}x}{x^{2}-a^{2}}}={\frac {1}{2a}}\ln \left|{\frac {x-a}{x+a}}\right|+C,\;a\not =0} ∫ d x x 2 ± a 2 = ln | x + x 2 ± a 2 | + C , a ≠ 0 {\displaystyle \int {\frac {{\mathsf {d}}x}{\sqrt {x^{2}\pm a^{2}}}}=\ln \left|x+{\sqrt {x^{2}\pm a^{2}}}\right|+C,\;a\not =0} ∫ a 2 − x 2 d x = x 2 a 2 − x 2 + a 2 2 arcsin x a + C {\displaystyle \int {\sqrt {a^{2}-x^{2}}}\;{\mathsf {d}}x={\frac {x}{2}}{\sqrt {a^{2}-x^{2}}}+{\frac {a^{2}}{2}}\arcsin {\frac {x}{a}}+C} ∫ x 2 ± a 2 d x = x 2 x 2 ± a 2 ± a 2 2 ln | x + x 2 ± a 2 | + C {\displaystyle \int {\sqrt {x^{2}\pm a^{2}}}\;{\mathsf {d}}x={\frac {x}{2}}{\sqrt {x^{2}\pm a^{2}}}\pm {\frac {a^{2}}{2}}\ln \left|x+{\sqrt {x^{2}\pm a^{2}}}\right|+C} ∫ a x + b d x = ( 2 b 3 a + 2 x 3 ) a x + b + C {\displaystyle \int {\sqrt {ax+b}}\;{\mathsf {d}}x=\left({2b \over 3a}+{2x \over 3}\right){\sqrt {ax+b}}+C} ∫ a x + b d x = 2 3 a ( a x + b ) 3 / 2 + C {\displaystyle \int {\sqrt {ax+b}}dx={2 \over 3a}(ax+b)^{3/2}+C} ∫ ( a + b ) 2 − x d x = − 2 ( a 2 + 2 a b + b 2 − x ) a 2 + 2 a b + b 2 − x 3 + C {\displaystyle \int {\sqrt {(a+b)^{2}-x}}dx=-{2(a^{2}+2ab+b^{2}-x){\sqrt {a^{2}+2ab+b^{2}-x}} \over 3}+C} ∫ a x 2 + b x d x = 2 b 2 a s i n ( a r c s e c ( 2 a x + b b ) ) + b 2 a c o s ( a r c s e c ( 2 a x + b b ) ) 2 ln ( | s i n ( a r c s e c ( 2 a x + b b ) ) − 1 s i n ( a r c s e c ( 2 a x + b b ) ) + 1 | ) 16 a 2 × c o s ( a r c s e c ( 2 a x + b b ) ) 2 + C {\displaystyle \int {\sqrt {ax^{2}+bx}}dx={{2b^{2}{\sqrt {a}}sin(arcsec({2ax+b \over b}))+b^{2}{\sqrt {a}}cos(arcsec({2ax+b \over b}))^{2}\ln(\left\vert {{sin(arcsec({{2ax+b} \over {b}}))-1} \over {sin(arcsec({{2ax+b} \over {b}}))+1}}\right\vert )} \over {16a^{2}\times cos(arcsec({{2ax+b} \over {b}}))^{2}}}+C}
∫ sin ( a x ) d x = − 1 a cos ( a x ) + C {\displaystyle \int \sin(ax)\;{\mathsf {d}}x=-{\frac {1}{a}}\cos(ax)+C} ∫ cos ( a x ) d x = 1 a sin ( a x ) + C {\displaystyle \int \cos(ax)\;{\mathsf {d}}x={\frac {1}{a}}\sin(ax)+C} ∫ tan x d x = − ln | cos x | + C {\displaystyle \int \tan x\;{\mathsf {d}}x=-\ln |\cos x|+C} ∫ c t g x d x = ln | sin x | + C {\displaystyle \int ctgx\;{\mathsf {d}}x=\ln |\sin x|+C} ∫ d x sin x = ln | tan x 2 | + C {\displaystyle \int {\frac {{\mathsf {d}}x}{\sin x}}=\ln \left|\tan {\frac {x}{2}}\right|+C} ∫ d x cos x = ln | tan ( x 2 + π 4 ) | + C {\displaystyle \int {\frac {{\mathsf {d}}x}{\cos x}}=\ln \left|\tan \left({\frac {x}{2}}+{\frac {\pi }{4}}\right)\right|+C} ∫ d x sin 2 x = − c t g x + C {\displaystyle \int {\frac {{\mathsf {d}}x}{\sin ^{2}x}}=-ctgx+C} ∫ d x cos 2 x = tan x + C {\displaystyle \int {\frac {{\mathsf {d}}x}{\cos ^{2}x}}=\tan x+C}
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