What is the integral of ((e^t)^2)/((e^{2t)+1)} ?

The integral of ((e^t)^2)/((e^{2t)+1)} is 1/2 ln|e^{2t}+1|+C

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integral of ((e^t)^2)/((e^{2t)+1)}

integral

\frac{( e ^{t} ) ^{2}}{( e ^{2 t} + 1 )}

\frac{1}{2} ln e^{2 t} + 1 + C

Solution steps

\int \frac{( e ^{t} ) ^{2}}{e ^{2 t} + 1} d t

(e^{t})^{2} = e^{2 t}

= \int \frac{e ^{2 t}}{e ^{2 t} + 1} d t

Apply u-substitution

: \int \frac{1}{2 u} d u

= \int \frac{1}{2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = e^{2 t} + 1

= \frac{1}{2} ln e^{2 t} + 1

Add a constant to the solution

= \frac{1}{2} ln e^{2 t} + 1 + C

\frac{x ^{2} + 1}{x ^{2} - 1}

\frac{1}{( 1 + x ) ^{0.5}}

2 sin (x) - x^{2} + 1

\frac{( x ^{2} + 4 )}{( x + 3 )}

f (x) = \frac{3 x + 3}{( 3 - 8 x + x ^{2} ) ^{\frac{1}{2}}}

What is the integral of ((e^t)^2)/((e^{2t)+1)} ?
The integral of ((e^t)^2)/((e^{2t)+1)} is 1/2 ln|e^{2t}+1|+C