Question

помогите найти неопределенный интеграл

Answer 1

$intlimits{frac{1}{(x-4)*ln^4(x-4)}} , dx= intlimits{frac{1}{ln^4(x-4)}}*frac{1}{x-4} , dx=$

$=intlimits{(ln(x-4))^{-4}}} , d(ln(x-4))=frac{(ln(x-4))^{-4+1}}{-4+1}+C=-frac{1}{3ln^3(x-4)}+C$
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$intlimits{frac{2x+5}{sqrt{5x^2+1}}},dx =intlimits { frac{2x}{ sqrt{5x^2+1} } } , dx +5 intlimits {frac{1}{ sqrt{5x^2+1} }} , dx =$

$=intlimits { frac{1}{ sqrt{5x^2+1} } } , d(x^2) +5 intlimits {frac{1}{ sqrt{5x^2+1} }} , dx =$

$= frac{1}{5} intlimits{(5x^2+1)^{-frac{1}{2}}},d(5x^2+1)+5* frac{1}{sqrt{5}}intlimits {frac{1}{sqrt{(sqrt{5}x)^2+1} }} , d(sqrt{5}x)=$

$= frac{1}{5}* frac{1}{-frac{1}{2}+1}*(5x^2+1)^{- frac{1}{2}+1}+sqrt{5}*ln|sqrt{5}x+ sqrt{5x^2+1} |+C=$

$= frac{2}{5}*sqrt{5x^2+1}+sqrt{5}*ln|sqrt{5}x+ sqrt{5x^2+1} |+C$
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$intlimits {x^3ln(x)} , dx=intlimits {ln(x)} , d(frac{x^4}{4})=frac{x^4}{4}*ln(x)-intlimits {frac{x^4}{4}} , d(ln(x))=$

$=frac{x^4}{4}*ln(x)-intlimits {frac{x^4}{4}}* frac{1}{x}, dx=$

$=frac{x^4}{4}*ln(x)-frac{1}{4}intlimits {x^3}, dx =frac{x^4}{4}*ln(x)-frac{1}{4}* frac{x^4}{4}+C=frac{x^4}{4}(ln(x)-frac{1}{4})+C$
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$intlimits { frac{2x^3-x^2+12x-2}{x^2-2x-8} } , dx$ = [деление в столбик] = 

$= intlimits {[2x+3+ frac{34x+22}{x^2-2x-8} ]} , dx=x^2+3x+ intlimits {frac{34x+22}{x^2-2x-8}} , dx$

$=x^2+3x+ intlimits {frac{34x+22}{(x-1)^2-3^2}} , dx =x^2+3x+ intlimits {frac{34x-34+56}{(x-1)^2-3^2}} , dx=$

$=x^2+3x+ 17intlimits {frac{2(x-1)}{(x-1)^2-3^2}} , dx+56intlimits {frac{1}{(x-1)^2-3^2}} , dx=$

$=x^2+3x+ 17intlimits {frac{1}{(x-1)^2-3^2}} , d((x-1)^2-3^2)+56intlimits {frac{1}{(x-1)^2-3^2}} , d(x-1)=$

$=x^2+3x+ 17ln|(x-1)^2-3^2|+56* frac{1}{2*3}*ln| frac{x-1-3}{x-1+3}|+C=$

$=x^2+3x+ 17*ln|x^2-2x-8|+frac{28}{3}ln| frac{x-4}{x+2}|+C$