What is the integral of 1/(2^x) ?

The integral of 1/(2^x) is -1/(ln(2)*2^x)+C

Popular >

integral of 1/(2^x)

\int \frac{1}{2 ^{x}} d x

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

Solution steps

\int \frac{1}{2 ^{x}} d x

Apply u-substitution

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

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

Apply the Power Rule

: - \frac{1}{u}

= \frac{1}{ln ( 2 )} (- \frac{1}{u})

Substitute back

u = 2^{x}

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

Simplify

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

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

Add a constant to the solution

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

y = x^{2} sin^{9} (x) + x cos^{- 4} (x)

\int \frac{1}{x ^{2} 4 x ^{2} + 9} d x

f (x) = 6 x^{3} - \frac{27}{2} x^{2} + 3

\frac{d}{d x} ((8 x + 5)^{2})

\frac{8}{7 x ^{2}}