Question

Integrate \[\int\limits \frac{e ^{2x}}{e ^{x}+3}dx\]

Answer 1

then?

Answer 2

this is better
let e^x+3=t

Answer 3

how can be the answer is \[e ^{x}+3\ln \left|e^{x}+3 \right|+C\]???

Answer 4

Numerator = (t-3)dt
Denominator= t

Answer 5

this method is Integration by substitution
Have U studied about it before?

Answer 6

yup.
Let => \[u = e ^{x}+3\]
\[du = e ^{x}\]

\[=\int\limits \frac{1}{u}\]
\[= \ln u + C\]
\[= \ln \left| e ^{x}+3 \right|+C\]

Answer 7

u droped one term!

Answer 8

what?

Answer 9

how?

Answer 10

i know i have a mistake

Answer 11

i dont know how to correct it.

Answer 12

well \[\frac{e^{2x}dx}{e^x+3}=\frac{e^{x}e^{x}dx}{e^x+3}\\ t=e^x+3 \\ dt=e^x dx\]
then u have
\[\frac{(t-3)dt}{t}\]

Answer 13

now try again

Answer 14

i can't get it. :( please show me.

Answer 15

is that ok till there?

Answer 16

are you typing the solution @mukushla ???

Answer 17

yes

Answer 18

\[\int\limits \frac{e^{2x}}{e^x+3}dx \\ \]
then we use a substitution

\[t=e^x+3 \\dt= e^x dx \\ e^x=t-3\]

now we have

\[\int\limits \frac{e^{2x}}{e^x+3}dx= \int\limits \frac{e^{x}e^{x}dx}{e^x+3}=\int\limits \frac{(t-3)}{t}dt=\int\limits (1-\frac{3}{t})dt=t-3 \ ln \left| t \right|+C\]

Answer 19

Thanks a lot @mukushla .

Answer 20

ur welcome