Question

MAA MATHEMATICAL ASSOCIATION OF AMERICA < webwork math-212-01-summer-2020 / assignment_5 / 4 Assignment 5: Problem 4 Previous Problem Problem List Next Problem (1 point) Use integration by parts and the technique of partial fractions to evaluate the integral /on(z? - 2 + 2) da Note: Use an upper-case "C" for the constant of integration Preview My Answers Submit Answers You have attempted this problem 5 times. Your overall recorded score is 0%

Answer 1

we have

9in(22 – 1 + 2)dac

integration by parts,

$u=ln(x^2-x+2) \Rightarrow du=\frac{2x-1}{x^2-x+2}$

and

v=9U9.7

we know that,

$\int uv'=uv-\int u'v$

we can say that,

91n (22 – 2 + 2)d.x = In (22 – 2 + 2). 92 2.0 – 1 . 9.cdr 22 – +2

$\int 9ln(x^2-x+2)dx=9xln(x^2-x+2)-9\int \frac{2x^2-x}{x^2-x+2}dx$

$\int 9ln(x^2-x+2)dx=9xln(x^2-x+2)-9\int \frac{2x^2-2x+4+x-4}{x^2-x+2}dx$

$\int 9ln(x^2-x+2)dx=9xln(x^2-x+2)-9\int \frac{2(x^2-x+2)+x-4}{x^2-x+2}dx$

$\int 9ln(x^2-x+2)dx=9xln(x^2-x+2)-9\int 2+\frac{x-4}{x^2-x+2}dx$

$\int 9ln(x^2-x+2)dx=9xln(x^2-x+2)-18x-9\int \frac{x-4}{(x-2)(x+1)}dx$

now,

$9\int \frac{x-4}{(x-2)(x+1)}dx=3\int \frac{3x-12}{(x-2)(x+1)}dx$

$9\int \frac{x-4}{(x-2)(x+1)}dx=3\int \frac{2x+2-5x+10}{(x-2)(x+1)}dx$

$9\int \frac{x-4}{(x-2)(x+1)}dx=3\int \frac{2(x+1)-5(x-2)}{(x-2)(x+1)}dx$

$9\int \frac{x-4}{(x-2)(x+1)}dx=3\int \frac{2}{x-2}-\frac{5}{x+1}dx$

$9\int \frac{x-4}{(x-2)(x+1)}dx=3\left [ 2ln|x-2|-5ln|x+1| \right ]+C$

put this value in equation 1),

$\int 9ln(x^2-x+2)dx=9xln(x^2-x+2)-18x-3\left [ 2ln|x-2|-5ln|x+1| \right ]+C$

$\int 9ln(x^2-x+2)dx=9xln(x^2-x+2)-18x-6ln|x-2|+15ln|x+1| \right ]+C$