# int (4x-2/(2x+3)^2) dx Find the indefinite integral

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...

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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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