Answered: "Evaluate the integral. ∫ex+e3xe2xdx"

IMLOG10ct

Answered question

2021-11-21

Evaluate the integral.

\int \frac{e^{x} + e^{3 x}}{e^{2 x}} d x

George Morin

Beginner2021-11-22Added 13 answers

We know that,

\int e^{a x} d x = \frac{e^{a x}}{a}

Evaluating the integral.

\int \frac{e^{x} + e^{3 x}}{e^{2 x}} d x = \int (\frac{e^{x}}{e^{2 x}} + \frac{e^{3 x}}{e^{2 x}}) d x

= \int (\frac{1}{e^{x}} + e^{x}) d x

= \int \frac{1}{e^{x}} d x + \int e^{x} d x

= \int e^{- x} d x + \int e^{x} d x

= \frac{e^{- x}}{- 1} + e^{x} + C

, where C is integration constant

= - e^{- x} + e^{x} + C

= e^{x} - e^{- x} + C

Therefore,

\int \frac{e^{x} + e^{3 x}}{e^{2 x}} d x = e^{x} - e^{- x} + C

Theirl1972

Beginner2021-11-23Added 22 answers

Step 1: Simplify

\frac{e^{x} + e^{3 x}}{e^{2 x}} \to \frac{1}{e^{x}} + e^{x}

\int \frac{1}{e^{x}} + e^{x} d x

Step 2: Use Sum Rule:

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

\int \frac{1}{e^{x}} d x + \int e^{x} d x

Step 3: Use Integration by Substitution on

\int \frac{1}{e^{x}} d x

.
Let

u = \frac{1}{e^{x}}, d u = - \frac{1}{e^{x}} d x, t h e n d x = - e^{\ln \frac{1}{u}} d u

Step 4: Using u and du above, rewrite

\int \frac{1}{e^{x}} d x

\int u \times - e^{\ln \frac{1}{u}} d u

Step 5: Use this rule:

\int a d x = a x + C

.
-u
Step 6: Substitute

u = \frac{1}{e^{x}}

back into the original integral.

- \frac{1}{e^{x}}

Step 7: Rewrite the integral with the completed substitution.

- \frac{1}{e^{x}} + \int e^{x} d x

Step 8: The integral of

e^{x} i s e^{x}

e^{x} - \frac{1}{e^{x}}

Step 9: Add constant.

e^{x} - \frac{1}{e^{x}} + C

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