What is the integral of sqrt(1+3^x) ?

The integral of sqrt(1+3^x) is -(ln|sqrt(1+3^x)+1|-ln|sqrt(1+3^x)-1|-2sqrt(1+3^x))/(ln(3))+C

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integral of sqrt(1+3^x)

\int 1 + 3^{x} d x

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

Solution steps

\int 1 + 3^{x} d x

Apply u-substitution

: \int \frac{u}{ln ( 3 ) ( u - 1 )} d u

= \int \frac{u}{ln ( 3 ) ( u - 1 )} d u

Take the constant out:

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

= \frac{1}{ln ( 3 )} \cdot \int \frac{u}{u - 1} d u

Apply u-substitution

: \int \frac{2 v ^{2}}{v ^{2} - 1} d v

= \frac{1}{ln ( 3 )} \cdot \int \frac{2 v ^{2}}{v ^{2} - 1} d v

Take the constant out:

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

= \frac{1}{ln ( 3 )} \cdot 2 \cdot \int \frac{v ^{2}}{v ^{2} - 1} d v

Factor

v^{2} - 1 : - (- v^{2} + 1)

= \frac{1}{ln ( 3 )} \cdot 2 \cdot \int \frac{v ^{2}}{- ( - v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{ln ( 3 )} \cdot 2 (- \int \frac{v ^{2}}{- v ^{2} + 1} d v)

\frac{v ^{2}}{- v ^{2} + 1} = \frac{1}{- v ^{2} + 1} - 1

= \frac{1}{ln ( 3 )} \cdot 2 (- \int \frac{1}{- v ^{2} + 1} - 1 d v)

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{ln ( 3 )} \cdot 2 (- (\int \frac{1}{- v ^{2} + 1} d v - \int 1 d v))

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

\int 1 d v = v

= \frac{1}{ln ( 3 )} \cdot 2 (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2} - v))

Substitute back

= \frac{1}{ln ( 3 )} \cdot 2 (- (\frac{ln 1 + 3 ^{x} + 1}{2} - \frac{ln 1 + 3 ^{x} - 1}{2} - 1 + 3^{x}))

Simplify

\frac{1}{ln ( 3 )} \cdot 2 (- (\frac{ln 1 + 3 ^{x} + 1}{2} - \frac{ln 1 + 3 ^{x} - 1}{2} - 1 + 3^{x})) : - \frac{ln 1 + 3 ^{x} + 1 - ln 1 + 3 ^{x} - 1 - 2 1 + 3 ^{x}}{ln ( 3 )}

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

Add a constant to the solution

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

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What is the integral of sqrt(1+3^x) ?
The integral of sqrt(1+3^x) is -(ln|sqrt(1+3^x)+1|-ln|sqrt(1+3^x)-1|-2sqrt(1+3^x))/(ln(3))+C