Calculus of a Single Variable, Chapter 8, 8.1, Section 8.1, Problem 20

int (4x - 2/(2x+3)^2)dx
To solve, express it as difference of two integrals.
= int 4x dx - int 2/(2x+3)^2dx
Then, apply negative exponent rule a^(-m)=1/a^m .
= int 4xdx - int 2(2x+3)^(-2)dx
For the second integral, apply the u-substitution method.

u = 2x + 3
du = 2dx

Expressing the second integral in terms of u variable, it becomes:
=int 4xdx - int (2x+3)^(-2) * 2dx
=int 4xdx - int u^(-2) du
For both integrals, apply the formula int x^ndx= x^(n+1)/(n+1)+C .
= (4x^2)/2 - u^(-1)/(-1) + C
=2x^2 + u^(-1) + C
= 2x^2 + 1/u + C
And, substitute back u = 2x + 3
=2x^2+1/(2x+3)+C

Therefore, int (4x - 2/(2x+3)^2)dx=2x^2+1/(2x+3)+C .

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