ludmilagena

Verified answer

1)   ∫sin²x cosx dx=∫sin²x d(sinx)=(1/3) * sin³x +C

2) ∫lnx dx=x*ln|x| -x + C

3) ∫xdx / cos²x² =(1/2)*∫ d(x²) /cos²x² =(1/2)*tgx² +C

4) A/x+B/(x-3) =(Ax-3A+Bx) /(x*(x-3))
    (x+2)/(x*(x-3))=(x(A+B)-3A)/(x*(x-3))
    -3A=2 --->  A=-2/3
     A+B=1 --->  B=5/3
∫(x+2)dx/(x*(x+2)) =(-2/3)*∫dx/x +(5/3)*∫dx/(x-3) =
 =(-2/3)*ln|x| +(5/3) *ln|x-3| +C