What is the integral of (2^t)/(2^t+3) ?

The integral of (2^t)/(2^t+3) is 1/(ln(2))ln|2^t+3|+C

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integral of (2^t)/(2^t+3)

\int \frac{2 ^{t}}{2 ^{t} + 3} d t

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

Solution steps

\int \frac{2 ^{t}}{2 ^{t} + 3} d t

Apply u-substitution

: \int \frac{1}{ln ( 2 ) u} d u

= \int \frac{1}{ln ( 2 ) u} d u

Take the constant out:

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

= \frac{1}{ln ( 2 )} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{ln ( 2 )} ln ∣ u ∣

Substitute back

u = 2^{t} + 3

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (7 e^{x^{8} y})

y^{^{'}} = 4 x - 6

x \to \infty lim (in x)

\frac{d}{d x} ((7 x^{2} + 6 x)^{7} (3 x - 1)^{4})

\frac{\partial}{\partial x} (x^{3} + y^{3} + x y)

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