Resolved: "Evaluate the integral. ∫e2x+e3xdx"

Carol Gates

Answered question

2021-11-06

Evaluate the integral.

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

Elberte

Skilled2021-11-07Added 95 answers

Step 1
Given integral is,

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

Step 2
To evaluate the given integral.
Given integral can be written as,

\int \sqrt{e^{2 x} + e^{3 x}} d x = \sqrt{e^{2 x} (1 + e^{x})} d x

= \int \sqrt{e^{2 x}} \sqrt{1 + e^{x}}

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

Apply the substitution,

1 + e^{x} = u

\Rightarrow d u = e^{x} d x

Then the integral becomes,

\int \sqrt{e^{2 x} + e^{3 x}} d x = \int \sqrt{u} d u

= \frac{u^{\frac{3}{2}}}{\frac{3}{2}} + c

Now substitute again

1 + e^{x} = u

we get,

\int \sqrt{e^{2 x} + e^{3 x}} d x = \frac{2}{3} {(1 + e^{x})}^{\frac{3}{2}} + c

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