Question

Integrate :-
$\displaystyle \int \dfrac{e^{2x}-2e^x}{e^{2x}+1}dx$

Answer 1

⟹ $\displaystyle \int \dfrac{e^{2x}-2e^x}{e^{2x}+1}dx$

⟹ $\displaystyle \int \dfrac{e^{2x}}{e^{2x}+1}dx - \displaystyle \int \dfrac{2e^{x}}{e^{2x}+1}dx$

⟹ $I1 = \displaystyle \int \dfrac{e^{2x}}{e^{2x}+1}dx$

⟹ $I2 = \displaystyle \int \dfrac{2e^{x}}{e^{2x}+1}dx$

First we will solve first part :-

⟹ $I1 = \displaystyle \int \dfrac{e^{2x}}{e^{2x}+1}dx$

we will solve this by using substitution method :-

⟹ ${e}^{2x} + 1 = t$

⟹ $2{e}^{2x} dx = dt$

⟹ $I1 = \frac{1}{2} \displaystyle \int \dfrac{dt}{t}$

⟹ $I1 = log(t) + c1$

Where c1 is constant

Now put the value of t :-

⟹ $I1 = log({e}^{2x} + 1) + c1$

Now we will solve the I2 :-

⟹ $I2 = \displaystyle \int \dfrac{2e^{x}}{e^{2x}+1}dx$

⟹ $let \: {e}^{x} = m$

⟹ ${e}^{x}dx = dm$

⟹ $I2 =2 \displaystyle \int \dfrac{dm}{m^{2}+1}$

⟹ $I2 =2 \ {tan}^{ - 1} (m) + c2$

put the value of m now :-

⟹ $I2 =2 \ {tan}^{ - 1} ( {e}^{x} ) + c2$

Therefore the final answer is :- I1 - I2

⟹ $\frac{1}{2} log({e}^{2x} + 1) - 2 \ {tan}^{ - 1} ( {e}^{x} ) + c$

[c = c1 + c2]

Answer :-

$\frac{1}{2} log({e}^{2x} + 1) - 2 \ {tan}^{ - 1} ( {e}^{x} ) + c$

Answer 2

GIVEN :-

$\begin{gathered}\begin{gathered}\begin{gathered}\\ \bullet\:{ \tt{ \int \dfrac{e^{2x}-2e^x}{e^{2x}+1}dx}}\\ \\ \end{gathered}\end{gathered}\end{gathered}$

TO FIND :-

$\begin{gathered}\begin{gathered}\\ \tt \bullet \: The \: Integrate\\ \\\end{gathered}\end{gathered}$

SOLUTION :-

$\begin{gathered}\begin{gathered}\begin{gathered}\\ : \implies{ \tt{ \int \dfrac{e^{2x}}{e^{2x}+1} - \frac{2 {e}^{x} }{ {e}^{2x} + 1} \: dx }}\\ \\ \end{gathered}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}\\ : \implies{ \tt{ \int \dfrac{e^{2x}}{e^{2x}+1} \: dx - \int\frac{2 {e}^{x} }{ {e}^{2x} + 1} \: dx }}\\ \\ \end{gathered}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}\\ : \implies{ \tt{ \frac{1}{2} \times in \bigg( {e}^{2x} + 1 \bigg) - \int\frac{2 {e}^{x} }{ {e}^{2x} + 1} \: dx }}\\ \\ \end{gathered}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}\\ : \implies{ \tt{ \frac{1}{2} \times in \bigg( {e}^{2x} + 1 \bigg) - 2 \: arctan \bigg( {e}^{x} \bigg) }}\\ \\ \end{gathered}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}\\ : \implies \large{ \boxed{ \tt{ \frac{1}{2} \times in \bigg( {e}^{2x} + 1 \bigg) - 2 \: arctan \bigg( {e}^{x} \bigg) + C}}} \\\end{gathered} \\ \\\end{gathered}\end{gathered}$