⟹
First we will solve first part :-
we will solve this by using substitution method :-
Where c1 is constant
Now put the value of t :-
Now we will solve the I2 :-
put the value of m now :-
Therefore the final answer is :- I1 - I2
[c = c1 + c2]
GIVEN :-
TO FIND :-
SOLUTION :-
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⟹![\displaystyle \int \dfrac{e^{2x}-2e^x}{e^{2x}+1}dx \displaystyle \int \dfrac{e^{2x}-2e^x}{e^{2x}+1}dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20%5Cdfrac%7Be%5E%7B2x%7D-2e%5Ex%7D%7Be%5E%7B2x%7D%2B1%7Ddx)
⟹![\displaystyle \int \dfrac{e^{2x}}{e^{2x}+1}dx - \displaystyle \int \dfrac{2e^{x}}{e^{2x}+1}dx \displaystyle \int \dfrac{e^{2x}}{e^{2x}+1}dx - \displaystyle \int \dfrac{2e^{x}}{e^{2x}+1}dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint%20%5Cdfrac%7Be%5E%7B2x%7D%7D%7Be%5E%7B2x%7D%2B1%7Ddx%20-%20%5Cdisplaystyle%20%5Cint%20%5Cdfrac%7B2e%5E%7Bx%7D%7D%7Be%5E%7B2x%7D%2B1%7Ddx)
⟹![I1 = \displaystyle \int \dfrac{e^{2x}}{e^{2x}+1}dx I1 = \displaystyle \int \dfrac{e^{2x}}{e^{2x}+1}dx](https://tex.z-dn.net/?f=I1%20%3D%20%5Cdisplaystyle%20%5Cint%20%5Cdfrac%7Be%5E%7B2x%7D%7D%7Be%5E%7B2x%7D%2B1%7Ddx)
⟹![I2 = \displaystyle \int \dfrac{2e^{x}}{e^{2x}+1}dx I2 = \displaystyle \int \dfrac{2e^{x}}{e^{2x}+1}dx](https://tex.z-dn.net/?f=I2%20%3D%20%5Cdisplaystyle%20%5Cint%20%5Cdfrac%7B2e%5E%7Bx%7D%7D%7Be%5E%7B2x%7D%2B1%7Ddx)
First we will solve first part :-
⟹![I1 = \displaystyle \int \dfrac{e^{2x}}{e^{2x}+1}dx I1 = \displaystyle \int \dfrac{e^{2x}}{e^{2x}+1}dx](https://tex.z-dn.net/?f=I1%20%3D%20%5Cdisplaystyle%20%5Cint%20%5Cdfrac%7Be%5E%7B2x%7D%7D%7Be%5E%7B2x%7D%2B1%7Ddx)
we will solve this by using substitution method :-
⟹![{e}^{2x} + 1 = t {e}^{2x} + 1 = t](https://tex.z-dn.net/?f=%20%7Be%7D%5E%7B2x%7D%20%20%2B%201%20%3D%20t)
⟹![2{e}^{2x} dx = dt 2{e}^{2x} dx = dt](https://tex.z-dn.net/?f=2%7Be%7D%5E%7B2x%7D%20dx%20%20%3D%20dt)
⟹![I1 = \frac{1}{2} \displaystyle \int \dfrac{dt}{t} I1 = \frac{1}{2} \displaystyle \int \dfrac{dt}{t}](https://tex.z-dn.net/?f=I1%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%5Cdisplaystyle%20%5Cint%20%5Cdfrac%7Bdt%7D%7Bt%7D)
⟹![I1 = log(t) + c1 I1 = log(t) + c1](https://tex.z-dn.net/?f=I1%20%3D%20%20log%28t%29%20%20%2B%20c1)
Where c1 is constant
Now put the value of t :-
⟹![I1 = log({e}^{2x} + 1) + c1 I1 = log({e}^{2x} + 1) + c1](https://tex.z-dn.net/?f=I1%20%3D%20%20log%28%7Be%7D%5E%7B2x%7D%20%20%2B%201%29%20%20%2B%20c1)
Now we will solve the I2 :-
⟹![I2 = \displaystyle \int \dfrac{2e^{x}}{e^{2x}+1}dx I2 = \displaystyle \int \dfrac{2e^{x}}{e^{2x}+1}dx](https://tex.z-dn.net/?f=I2%20%3D%20%5Cdisplaystyle%20%5Cint%20%5Cdfrac%7B2e%5E%7Bx%7D%7D%7Be%5E%7B2x%7D%2B1%7Ddx)
⟹![let \: {e}^{x} = m let \: {e}^{x} = m](https://tex.z-dn.net/?f=let%20%5C%3A%20%20%7Be%7D%5E%7Bx%7D%20%20%3D%20m)
⟹![{e}^{x}dx = dm {e}^{x}dx = dm](https://tex.z-dn.net/?f=%20%7Be%7D%5E%7Bx%7Ddx%20%20%3D%20dm)
⟹![I2 =2 \displaystyle \int \dfrac{dm}{m^{2}+1} I2 =2 \displaystyle \int \dfrac{dm}{m^{2}+1}](https://tex.z-dn.net/?f=I2%20%3D2%20%5Cdisplaystyle%20%5Cint%20%5Cdfrac%7Bdm%7D%7Bm%5E%7B2%7D%2B1%7D)
⟹![I2 =2 \ {tan}^{ - 1} (m) + c2 I2 =2 \ {tan}^{ - 1} (m) + c2](https://tex.z-dn.net/?f=I2%20%3D2%20%20%20%5C%20%7Btan%7D%5E%7B%20-%201%7D%20%28m%29%20%20%2B%20c2)
put the value of m now :-
⟹![I2 =2 \ {tan}^{ - 1} ( {e}^{x} ) + c2 I2 =2 \ {tan}^{ - 1} ( {e}^{x} ) + c2](https://tex.z-dn.net/?f=I2%20%3D2%20%20%20%5C%20%7Btan%7D%5E%7B%20-%201%7D%20%28%20%7Be%7D%5E%7Bx%7D%20%29%20%20%2B%20c2)
Therefore the final answer is :- I1 - I2
⟹![\frac{1}{2} log({e}^{2x} + 1) - 2 \ {tan}^{ - 1} ( {e}^{x} ) + c \frac{1}{2} log({e}^{2x} + 1) - 2 \ {tan}^{ - 1} ( {e}^{x} ) + c](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B2%7D%20log%28%7Be%7D%5E%7B2x%7D%20%20%2B%201%29%20%20%20-%202%20%20%20%5C%20%7Btan%7D%5E%7B%20-%201%7D%20%28%20%7Be%7D%5E%7Bx%7D%20%29%20%20%20%2B%20c)
[c = c1 + c2]
Answer :-
GIVEN :-
TO FIND :-
SOLUTION :-