What is the integral of (e^tsqrt(e^t+2))/(e^t+6) ?

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

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integral of (e^tsqrt(e^t+2))/(e^t+6)

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

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

Solution steps

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

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}{r ^{2} + 1} + 1) d r

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}{r ^{2} + 1} + 1 d r

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}{r ^{2} + 1} d r + \int 1 d r)

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

\int 1 d r = r

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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