Question

integrate (dx)/(3 + 4e ^ x) dx​

Answer 1

Given to find,

$\small\displaystyle\longrightarrow I=\int\dfrac{dx}{3+4e^x}$

Multiplying and dividing by $\small4e^x,$

$\small\displaystyle\longrightarrow I=\int\dfrac{4e^x\ dx}{4e^x(3+4e^x)}\quad\dots(1)$

Substitute,

$\small\longrightarrow u=3+4e^x$

Then we get,

$\small\longrightarrow4e^x=u-3$

$\small\longrightarrow du=4e^x\ dx$

Then (1) becomes,

$\small\displaystyle\longrightarrow I=\int\dfrac{du}{(u-3)u}$

Multiplying and dividing by 3,

$\small\displaystyle\longrightarrow I=\int\dfrac{3\ du}{3(u-3)u}$

$\small\displaystyle\longrightarrow I=\dfrac{1}{3}\int\dfrac{u-(u-3)\ du}{(u-3)u}$

$\small\displaystyle\longrightarrow I=\dfrac{1}{3}\int\left(\dfrac{1}{u-3}-\dfrac{1}{u}\right)\ du$

$\small\displaystyle\longrightarrow I=\dfrac{1}{3}\left(\int\dfrac{du}{u-3}-\int\dfrac{du}{u}\right)$

$\small\displaystyle\longrightarrow I=\dfrac{1}{3}\big(\log|u-3|-\log|u|\big)+C_1$

Undoing substitution $\smallu=3+4e^x,$

$\small\displaystyle\longrightarrow I=\dfrac{1}{3}\big(\log|4e^x|-\log|3+4e^x|\big)+C_1$

$\small\displaystyle\longrightarrow I=\dfrac{1}{3}\big(\log4+x-\log|3+4e^x|\big)+C_1$

$\small\displaystyle\longrightarrow I=\dfrac{1}{3}\log4+\dfrac{1}{3}\big(x-\log|3+4e^x|\big)+C_1$

Taking $\small\dfrac{1}{3}\log4+C_1=C,$

$\small\text{ \displaystyle\longrightarrow\underline{\underline{I=\dfrac{1}{3}\big(x-\log|3+4e^x|\big)+C}} }$

Answer 2

${\orange{ \boxed{ \boxed{ \begin{array}{cc} \sf \: let \: \\ \\ I = \displaystyle \int \: \rm \: \frac{1}{3 + 4 {e}^{x} } \: \: dx \\ \\ {\pink{{ \boxed{ \begin{array}{cc} \sf \: substitute \: \\ \rm \: u = 3 + 4 {e}^{x} \\ \rm \implies \: \frac{du}{dx} = 4 {e}^{x} \\ \rm \implies \:dx = \frac{1}{4 {e}^{x} }du \end{array}}}}} \\ \\ = \displaystyle \int \: \rm \: \frac{1}{4 {e}^{x} \: .u } \: du \\ \\ = \displaystyle \int \: \rm \: \frac{1}{(u - 3)u} \: du \\ \\ = \displaystyle \int \: \rm \: \frac{1}{(1 - \frac{3}{u} ) {u}^{2} } \: du \\ \\ {\pink{ { \boxed{ \begin{array}{cc} \sf \: substitute \: \\ \rm \: v = 1 - \frac{3}{u} \\ \\ \rm \implies \: \frac{dv}{du} = \frac{3}{ {u}^{2} } \\ \\ \rm \implies \:du = \frac{ {u}^{2} }{3} \: dv \end{array}}}}} \\ \\ = \frac{1}{3} \displaystyle \int \: \rm \: \frac{1}{v} \: dv \\ \\ = \frac{1}{3}\times \rm \: \ln \: v + c \\ \\ \rm = \frac{1}{3} \times ln(1 - \frac{3}{u} ) + c \\ \\ \rm = \frac{1}{3} \times \ln(1 - \frac{3}{3 + 4 {e}^{x} } ) + c \\ \\ \sf = answer \end{array}}}}}$

Note: C is a integral constant