int (2x^2+7x-3)/(x-2)dx
To solve, divide the numerator by the denominator (see attached figure).
= int (2x + 11 + 19/(x-2)) dx
Express it as sum of three integrals.
= int 2xdx + int11dx + int 19/(x-2)dx
For the first integral, apply the formula int x^ndx = x^(n+1)/(n+1)+C .
For the second integral, apply the formula int adx = ax + C .
= (2x^2)/2 + 11x + C + int 19/(x-2)dx
=x^2+11x+C + int 19/(x-2)dx
For the third integral, use u-substitution method.
Let,
u = x - 2
Differentiate u.
du = dx
Then, plug-in them to the third integral.
=x^2+11x+C+19int 1/(x-2)dx
=x^2+11x+C+19int 1/udu
To take the integral of it, apply the formula int 1/xdx =ln|x| +C .
= x^2+11x + 19ln|u| + C
And substitute back u = x-2 .
=x^2+11x+19ln|x-2|+C
Therefore, int (2x^2+7x-3)/(x-2)dx = x^2+11x + 19ln|x-2| + C .
Friday, November 29, 2019
Calculus of a Single Variable, Chapter 5, 5.2, Section 5.2, Problem 16
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