MHB 4.2.1 AP calc exam another int with e

[karush]

Evaluate $\displaystyle\int\dfrac{e^{2x}}{1+e^x} \, dx=$

$a.\quad \tan^{-1}e^x+C$
$b.\quad 1+e^x-\ln(1+e^1)+C$
$c.\quad x-x+\ln |1+e^x|+C$
$d.\quad e^x+\frac{1}{(e^x+1)^2}+C$
$e.\quad {none}$

ok I was going to use $u=1+e^x\quad du=e^x dx$ but maybe not best

btw I tried to use array on the choices but its was all underlined in preview

 

[skeeter]

l$u=e^x \implies du = e^x \, dx$

$\displaystyle \int \dfrac{u}{1+u} \, du$

$\displaystyle \int \dfrac{u+1-1}{1+u} \, du$

$\displaystyle \int 1 - \dfrac{1}{1+u} \, du$

$u - \ln(1+u) + C$

$e^x - \ln(1+e^x) + C$

btw ... AP exams never use “none” as a choice

 

[karush]

yeah I know but typically they have 5 not 4 choices Ill invent some bogus answer

that is a very rare way to solve it not sure if I even saw that unless you are completing the square

 

[skeeter]

that is a very rare way to solve it not sure if I even saw that unless you are completing the square

?

Evaluate $\displaystyle\int\dfrac{e^{2x}}{1+e^x} \, dx=$

ok I was going to use $u=1+e^x\quad du=e^x dx$ but maybe not best

that works, too ...

$u = 1+e^x \implies du = e^{x} \, dx \text{ and } e^x = u-1$

$$\int \dfrac{u-1}{u} \, du$$

$$\int 1 - \dfrac{1}{u} \, du$$

$u - \ln(u) + C$

$1+e^x - \ln(1+e^x) + C$

 

[HOI]

l$u=e^x \implies du = e^x \, dx$

$\displaystyle \int \dfrac{u}{1+u} \, du$

The given problem had $e^{2x}= (e^x)^2$ in the denominator so the integral is
$\int \dfrac{u^2}{1+ u}{du}= \int 1- \dfrac{1}{1+ u} du$$= u- ln|1+ u|+ C= e^x- ln|1+ e^x|+ C$

$\displaystyle \int \dfrac{u+1-1}{1+u} \, du$

$\displaystyle \int 1 - \dfrac{1}{1+u} \, du$

$u - \ln(1+u) + C$

$e^x - \ln(1+e^x) + C$

btw ... AP exams never use “none” as a choice

 

[skeeter]

The given problem had $e^{2x}= (e^x)^2$ in the denominator so the integral is
$\int \dfrac{u^2}{1+ u}{du}= \int 1- \dfrac{1}{1+ u} du$$= u- ln|1+ u|+ C= e^x- ln|1+ e^x|+ C$

huh?

Evaluate $\displaystyle \int\dfrac{e^{2x}}{{\color{red}1+e^x}} \, dx=$

$a.\quad \tan^{-1}e^x+C$
$b.\quad 1+e^x-\ln(1+e^1)+C$
$c.\quad x-x+\ln |1+e^x|+C$
$d.\quad e^x+\frac{1}{(e^x+1)^2}+C$
$e.\quad {none}$

ok I was going to use $u=1+e^x\quad du=e^x dx$ but maybe not best

btw I tried to use array on the choices but its was all underlined in preview