6. Find dy/dx. y= ln (2x/(x+3))

ln 2x - ln x+3

1/2x - 1/x+3

(x+3)/2x(x+3) - (2x)/2x(x+3)

= -x+3/2x(x+3)

Answer is: 3/x(x+3)

7. Find dy/dx. y= ln (x^3 + 3x)^3

3ln(x^3+3x)^2 (3x^2+3)

= (9x^2+9)ln(3x^2+3)^2

Answer is: 9(x^2 +1)/x(x^2+3)

8. Find the derivative. f(x) = ln √x^2 +1) / x(2x^3 -1)^2 Note: radical is only over the numerator.

this is what i have so far but i'm not getting the answer:

ln (x^2 +1)^1/2 - ln x(2x^3 -2)^2

1/2 ln(x^2+1) - 2 ln x(2x^3-1)

1/2 n(x^2+1) - 2 ln x + ln (2x^3-1)

dy/dx = 1/2 x (1/x^2+1)2x - (2 (1/x) + (1/2x^3-1)6x^2)

= (x/x^2 +1) - (2/x) + (6x^2/2x^3-1)

but the answer is: (x/x^2 +1) - (1/x) - (12x^2/2x^3-1)

14. ∫ (2x^2 + x -3) / (x-2) dx

I did long division and got 2x-3-(3/x-2)

∫2x-3-(3/x-2)dx

2x^2/2 - 3x - (3∫1/x-2)dx

=x^2 - 3x - 3ln |x-2| +c

Answer is: x^2 + 5x + 7ln |x-2| +c

22. ∫ (sinx^2 - cosx^2) / sinx dx

Answer is: -2cosx+ln |cscx + cotx| +c

16. ∫ ln√x)) /x dx

∫ln (x)^1/2 / x

∫1/2 ln (x) / x

Let u= lnx

du= 1/x dx

1/2 ∫u du

Answer is: 1/4 (ln x)^2 +c