Using the substitution $t=tan(\frac{x}{2})$ with ${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{dx}{sinx+tanx}=2ln|sec(\frac{x}{2})|+\frac{1}{2}se{c}^2(\frac{x}{2})+C\frac{1}{2}ln|sin(\frac{x}{2})|-si{n}^2(\frac{x}{2})+C\frac{1}{2}ln|tan(\frac{x}{2})|-\frac{1}{4}ta{n}^2(\frac{x}{2})+C\frac{3}{4}ln|tan(\frac{x}{2})|+\frac{1}{2}ta{n}^2(\frac{x}{2})+C\frac{1}{2}ln|tan(\frac{x}{2})|-ta{n}^2(\frac{x}{2})+C-,$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{dx}{sinx+tanx}=$

$A.2ln|sec(\frac{x}{2})|+\frac{1}{2}se{c}^2(\frac{x}{2})+C$

$B.\frac{1}{2}ln|sin(\frac{x}{2})|-si{n}^2(\frac{x}{2})+C$

$C.\frac{1}{2}ln|tan(\frac{x}{2})|-\frac{1}{4}ta{n}^2(\frac{x}{2})+C$

$D.\frac{3}{4}ln|tan(\frac{x}{2})|+\frac{1}{2}ta{n}^2(\frac{x}{2})+C$

$E.\frac{1}{2}ln|tan(\frac{x}{2})|-ta{n}^2(\frac{x}{2})+C$