Question: "Evaluate the following integrals. ∫exe2x+2ex+17dx"

bobbie71G

Answered question

2021-10-20

Evaluate the following integrals.

\int \frac{e^{x}}{e^{2 x} + 2 e^{x} + 17} d x

jlo2niT

Skilled2021-10-21Added 96 answers

Given that:
The integral is

\int \frac{e^{x}}{e^{2 x} + 2 e^{x} + 17} d x

By using the formula,

\int \frac{1}{x^{2} + a^{2}} d x = \frac{1}{a} \arctan (\frac{x}{a}) + C

By using substitution,
Substitute

v = e^{x}

, then

d v = e^{x} d x

Then,

\int \frac{e^{x}}{e^{2 x} + 2 e^{x} + 17} d x = \int \frac{1}{v^{2} + 2 v + 17} d v

= \int \frac{1}{v^{2} + 2 v + 1 - 1 + 17} d v

= \int \frac{1}{(v^{2} + 2 v + 1) + 16} d v

= \int \frac{1}{{(v + 1)}^{2} + 16} d v

Substitute u=v+1, du=dv
Then,

= \int \frac{1}{u^{2} + 16} d u

= \int \frac{1}{u^{2} + 4^{2}} d u

= \frac{1}{4} \arctan (\frac{u}{4})

Substitute back

u = v + 1

and

v = e^{x}

, then

u = e^{x} + 1

\int \frac{e^{x}}{e^{2 x} + 2 e^{x} + 17} d x = \frac{1}{4} \arctan (\frac{u}{4})

= \frac{1}{4} \arctan (\frac{e^{x} + 1}{4}) + C

Where C is the integration constant.
Therefore,

\int \frac{e^{x}}{e^{2 x} + 2 e^{x} + 17} d x = \frac{1}{4} \arctan (\frac{e^{x} + 1}{4}) + C

, where c is the integration constant.

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