Question

The value of $\int \frac{dx}{3-2x-x^2}$ will be

• A. $\frac{1}{4}log\left(\frac{3+x}{1-x}\right)$
• B. $\frac{1}{3}log\left(\frac{3+x}{1-x}\right)$
• C. $\frac{1}{2}log\left(\frac{3+x}{1-x}\right)$
• D. $log\left(\frac{1-x}{3+x}\right)$

Answer 1

The correct option is A $\frac{1}{4}log\left(\frac{3+x}{1-x}\right)$

$\int \frac{dx}{3-2x-x^2}=\int \frac{dx}{4-(x^2+2x+1)}=\int \frac{dx}{4-(x+1{)}^2}=\int \frac{dt}{(2{)}^2-t^2}$
Where $x+1=t,\therefore dx=dt\therefore I=\frac{1}{2.2}log\left(\frac{2+t}{2-t}\right)=\frac{1}{4}log\left(\frac{2+x+1}{2-x-1}\right)=\frac{1}{4}log\left(\frac{3+x}{1-x}\right)$