#\int(x^3+x)/(x-1)dx#?

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Long division got me #\intx^2+x+2+2/(x-1)dx#

Solving:
#\int2/(x-1)dx# by u-sub (#u=x-1# and #du=dx#) becomes #2\int1/udu=2\lnu+C=2\ln|x-1|+C#
#\therefore\intx^2+x+2+2/(x-1)dx=x^3/3+x^2/2+2x+2\ln|x-1|+C#

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Answer 1 · 2018-04-02T18:37:13

#I=x^3/3+x^2/2+2x+2ln|x-1|+c#
i.e.Your answer is O.K.

Here,

#I=int(x^3+x)/(x-1)dx=int(x^3-1+(x-1)+2)/(x-1)dx#

#=int((x-1)(x^2+x+1)+(x-1)+2)/(x-1)dx#

#=int((x^2+x+1)+1)dx+int2/(x-1)dx#

#=x^3/3+x^2/2+x+x+2ln|x-1|+c#

#=x^3/3+x^2/2+2x+2ln|x-1|+c#