What is the integral of 1/(\sqrt[5]{e^{2t)}} ?

The integral of 1/(\sqrt[5]{e^{2t)}} is -5/2 e^{-(2t)/5}+C

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integral of 1/(\sqrt[5]{e^{2t)}}

- \frac{5}{2} e^{- \frac{2 t}{5}} + C

Solution steps

= \int \frac{1 - e ^{\frac{2 t}{5}}}{e ^{\frac{2 t}{5}}} + 1 d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1 - e ^{\frac{2 t}{5}}}{e ^{\frac{2 t}{5}}} d t + \int 1 d t

\int \frac{1 - e ^{\frac{2 t}{5}}}{e ^{\frac{2 t}{5}}} d t = - \frac{5}{2} e^{- \frac{2 t}{5}} - t

\int 1 d t = t

= - \frac{5}{2} e^{- \frac{2 t}{5}} - t + t

Simplify

= - \frac{5}{2} e^{- \frac{2 t}{5}}

Add a constant to the solution

= - \frac{5}{2} e^{- \frac{2 t}{5}} + C

\int (7 + e) d x

\int_{1}^{\infty} \frac{ln ( x )}{x} d x

\frac{d}{d x} (x^{4} - \frac{1}{x})

x \to 4 lim (x^{2} - 4 x + 1)

f (x) = 3 x - 5

What is the integral of 1/(\sqrt[5]{e^{2t)}} ?
The integral of 1/(\sqrt[5]{e^{2t)}} is -5/2 e^{-(2t)/5}+C