What is the integral of sqrt(e^t-3) ?

The integral of sqrt(e^t-3) is 2sqrt(3)(-arctan(sqrt(1/3 (e^t-3)))+sqrt(1/3 (e^t-3)))+C

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integral of sqrt(e^t-3)

\int e^{t} - 3 d t

23 (- arctan (\frac{1}{3} (e^{t} - 3)) + \frac{1}{3} (e^{t} - 3)) + C

Solution steps

\int e^{t} - 3 d t

Apply u-substitution

: \int \frac{u - 3}{u} d u

= \int \frac{u - 3}{u} d u

Apply u-substitution

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

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

\frac{v ^{2}}{v ^{2} + 3} = - \frac{3}{v ^{2} + 3} + 1

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

Apply Integral Substitution

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

Take the constant out:

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

= 23 \cdot \int - \frac{1}{w ^{2} + 1} + 1 d w

Apply the Sum Rule:

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

= 23 (- \int \frac{1}{w ^{2} + 1} d w + \int 1 d w)

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

\int 1 d w = w

= 23 (- arctan (w) + w)

Substitute back

= 23 (- arctan (\frac{e ^{t} - 3}{3}) + \frac{e ^{t} - 3}{3})

Simplify

23 (- arctan (\frac{e ^{t} - 3}{3}) + \frac{e ^{t} - 3}{3}) : 23 (- arctan (\frac{1}{3} (e^{t} - 3)) + \frac{1}{3} (e^{t} - 3))

= 23 (- arctan (\frac{1}{3} (e^{t} - 3)) + \frac{1}{3} (e^{t} - 3))

Add a constant to the solution

= 23 (- arctan (\frac{1}{3} (e^{t} - 3)) + \frac{1}{3} (e^{t} - 3)) + C

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What is the integral of sqrt(e^t-3) ?
The integral of sqrt(e^t-3) is 2sqrt(3)(-arctan(sqrt(1/3 (e^t-3)))+sqrt(1/3 (e^t-3)))+C