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Popular >

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

Solution

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

Solution

ln \frac{2}{3} e^{x} + \frac{1}{3} + \frac{1}{3} (4 e^{x} + 4 e^{2 x} + 4) + C

Solution steps

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

Apply u-substitution

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

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

Complete the square

1 + u + u^{2} : (u + \frac{1}{2})^{2} + \frac{3}{4}

= \int \frac{1}{( u + \frac{1}{2} ) ^{2} + \frac{3}{4}} d u

Apply u-substitution

: \int \frac{2}{4 v ^{2} + 3} d v

= \int \frac{2}{4 v ^{2} + 3} d v

Take the constant out:

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

= 2 \cdot \int \frac{1}{4 v ^{2} + 3} d v

Apply Trigonometric Substitution

: \int \frac{sec ( w )}{2} d w

= 2 \cdot \int \frac{sec ( w )}{2} d w

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int sec (w) d w

Use the common integral:

\int sec (w) d w = ln ∣ tan (w) + sec (w) ∣

= 2 \cdot \frac{1}{2} ln ∣ tan (w) + sec (w) ∣

Substitute back

= 2 \cdot \frac{1}{2} ln tan (arctan (\frac{2}{3} (e^{x} + \frac{1}{2}))) + sec (arctan (\frac{2}{3} (e^{x} + \frac{1}{2})))

Simplify

2 \cdot \frac{1}{2} ln tan (arctan (\frac{2}{3} (e^{x} + \frac{1}{2}))) + sec (arctan (\frac{2}{3} (e^{x} + \frac{1}{2}))) : ln \frac{2}{3} e^{x} + \frac{1}{3} + \frac{1}{3} (4 e^{x} + 4 e^{2 x} + 4)

= ln \frac{2}{3} e^{x} + \frac{1}{3} + \frac{1}{3} (4 e^{x} + 4 e^{2 x} + 4)

Add a constant to the solution

= ln \frac{2}{3} e^{x} + \frac{1}{3} + \frac{1}{3} (4 e^{x} + 4 e^{2 x} + 4) + C

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