Question

lf $\int \frac{dx}{a{e}^{mx}+b{e}^{-mx}}=K{\tan }^{-1}\left(P{e}^{mx}\right)+C$, then $K,P=$

• A. $K=\frac{1}{\sqrt{ab}},P=\sqrt{\frac{a}{b}}$
• B. $K=\frac{1}{m\sqrt{ab}},P=\sqrt{\frac{a}{b}}$
• C. $K=m\sqrt{ab},P=\sqrt{\frac{b}{a}}$
• D. $K=\frac{1}{m\sqrt{ab}},P=\sqrt{\frac{b}{a}}$

Answer 1

The correct option is B $K=\frac{1}{m\sqrt{ab}},P=\sqrt{\frac{a}{b}}$

$\int \frac{e^{mx}dx}{a{e}^{2mx}+b}$
$e^{mx}=t$
$m{e}^{mx}dx=dt$
$\int \frac{dx}{m(a{t}^2+b)}=\frac{1}{m}\frac{1}{\sqrt{b}}\cdot \frac{1}{\sqrt{a}}{\tan }^{-1}\left(\frac{\sqrt{a}t}{\sqrt{b}}\right)+c$
$=\frac{1}{m\sqrt{ab}}{\tan }^{-1}\left(\sqrt{\frac{a}{b}t}\right)+c$
$k=\frac{1}{m\sqrt{ab}},p=\sqrt{\frac{a}{b}}$