True or False:
Consider the integral ∫x3+xx−1dx\displaystyle{\int \frac{x^3+x}{x-1}}dx∫x−1x3+xdx To evaluate the integral, we need to first perform the long division.
True
x3+xx−1=x2+x2+xx−1=x2+x+2+2x−1x3+xx−1=x2−x+2−2x+1Apply the Sum Rule:∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx=x33−x22+2x−2ln∣x+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|x−1x3+x=x2+x−1x2+x=x2+x+2+x−12x−1x3+x=x2−x+2−x+12ApplytheSumRule:∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx=3x3−2x2+2x−2ln∣x+1∣
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments