Question #233059

True or False:

 Consider the integral x3+xx1dx\displaystyle{\int \frac{x^3+x}{x-1}}dx To evaluate the integral, we need to first perform the long division.


1
Expert's answer
2021-09-08T10:14:31-0400

True


x3+xx1=x2+x2+xx1=x2+x+2+2x1x3+xx1=x2x+22x+1ApplytheSumRule:f(x)±g(x)dx=f(x)dx±g(x)dx=x33x22+2x2lnx+1\frac{x^3+x}{x-1}=x^2+\frac{x^2+x}{x-1}\\ =x^2+x+2+\frac{2}{x-1}\\ \frac{x^3+x}{x-1}=x^2-x+2-\frac{2}{x+1}\\ \mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\ =\frac{x^3}{3}-\frac{x^2}{2}+2x-2\ln \left|x+1\right|


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