What is the integral of (e^xsqrt(e^x+2))/(e^x+6) ?

The integral of (e^xsqrt(e^x+2))/(e^x+6) is 4(-arctan(1/2 sqrt(e^x+2))+1/2 sqrt(e^x+2))+C

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integral of (e^xsqrt(e^x+2))/(e^x+6)

\int \frac{e ^{x} e ^{x} + 2}{e ^{x} + 6} d x

4 (- arctan (\frac{1}{2} e^{x} + 2) + \frac{1}{2} e^{x} + 2) + C

Solution steps

\int \frac{e ^{x} e ^{x} + 2}{e ^{x} + 6} d x

Apply u-substitution

: \int \frac{u + 2}{u + 6} d u

= \int \frac{u + 2}{u + 6} d u

Apply u-substitution

: \int \frac{v}{v + 4} d v

= \int \frac{v}{v + 4} d v

Apply u-substitution

: \int \frac{2 w ^{2}}{w ^{2} + 4} d w

= \int \frac{2 w ^{2}}{w ^{2} + 4} d w

Take the constant out:

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

= 2 \cdot \int \frac{w ^{2}}{w ^{2} + 4} d w

\frac{w ^{2}}{w ^{2} + 4} = - \frac{4}{w ^{2} + 4} + 1

= 2 \cdot \int - \frac{4}{w ^{2} + 4} + 1 d w

Apply Integral Substitution

= 2 \cdot \int 2 (- \frac{1}{t ^{2} + 1} + 1) d t

Take the constant out:

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

= 2 \cdot 2 \cdot \int - \frac{1}{t ^{2} + 1} + 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

= 2 \cdot 2 (- \int \frac{1}{t ^{2} + 1} d t + \int 1 d t)

\int \frac{1}{t ^{2} + 1} d t = arctan (t)

\int 1 d t = t

= 2 \cdot 2 (- arctan (t) + t)

Substitute back

= 2 \cdot 2 (- arctan (\frac{e ^{x} + 2}{2}) + \frac{e ^{x} + 2}{2})

Simplify

2 \cdot 2 (- arctan (\frac{e ^{x} + 2}{2}) + \frac{e ^{x} + 2}{2}) : 4 (- arctan (\frac{1}{2} e^{x} + 2) + \frac{1}{2} e^{x} + 2)

= 4 (- arctan (\frac{1}{2} e^{x} + 2) + \frac{1}{2} e^{x} + 2)

Add a constant to the solution

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

\frac{d}{d x} (ln (e^{5 x}))

\int sin^{5} (\frac{x}{7}) cos (\frac{x}{7}) d x

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

\frac{d ^{2}}{d x ^{2}} (\frac{( x - 3 ) ( x ^{2} + 3 x + 9 )}{x ^{3}})

\int - \frac{1}{5} cos (5 x) d x

What is the integral of (e^xsqrt(e^x+2))/(e^x+6) ?
The integral of (e^xsqrt(e^x+2))/(e^x+6) is 4(-arctan(1/2 sqrt(e^x+2))+1/2 sqrt(e^x+2))+C