Answer to Question #174632 in Calculus for Phyroe

Question #174632

Evaluate the integral of (x³- 2x+5)/(x-3) dx


1
Expert's answer
2021-03-31T08:47:23-0400

Convert the expression x32x+5x3\frac{x^3-2x+5}{x-3} so that the numerator contains a polynomial of smaller degree than the denominator

x32x+5x3=x33x2+3x29x+7x21+26x3=\frac{x^3-2x+5}{x-3}=\frac{x^3-3x^2+3x^2-9x+7x-21+26}{x-3}=


=x2(x3)+3x(x3)+7(x3)+26x3=x2+3x+7+26x3=\frac{x^2(x-3)+3x(x-3)+7(x-3)+26}{x-3}=x^2+3x+7+\frac{26}{x-3}

Therefore, we can start evaluating the integral,


x32x+5x3dx=(x2+3x+7+26x3)dx=\intop\frac{x^3-2x+5}{x-3}dx=\intop(x^2+3x+7+\frac{26}{x-3})dx=

=x33+3x22+7x+26lnx3+C=\frac{x^3}{3}+\frac{3x^2}{2}+7x+26 ln|x-3|+C


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