What is the integral of 2^{1-2x} ?

The integral of 2^{1-2x} is -1/(ln(2)*4^x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 2^{1-2x}

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

- \frac{1}{ln ( 2 ) \cdot 4 ^{x}} + C

Solution steps

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

Apply u-substitution

: \int - 2^{u - 1} d u

= \int - 2^{u - 1} d u

Take the constant out:

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

= - \int 2^{u - 1} d u

Simplify

2^{u - 1} : 2^{u} \cdot 2^{- 1}

= - \int 2^{u} \cdot 2^{- 1} d u

Take the constant out:

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

= - 2^{- 1} \cdot \int 2^{u} d u

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= - \frac{1}{2} \cdot \int 2^{u} d u

\int a^{x} d x = \frac{a ^{x}}{ln a}

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

Substitute back

u = 1 - 2 x

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

Simplify

- \frac{1}{2} \cdot \frac{2 ^{1 - 2 x}}{ln ( 2 )} : - \frac{1}{ln ( 2 ) \cdot 4 ^{x}}

= - \frac{1}{ln ( 2 ) \cdot 4 ^{x}}

Add a constant to the solution

= - \frac{1}{ln ( 2 ) \cdot 4 ^{x}} + C

\int sec^{4} (θ) tan^{2} (θ) d θ

\int \frac{x}{( 1 + cos ( x ) )} d x

\int e^{(x^{2})^{2}} d (x^{2})^{2}

\int \frac{5 x}{1 + x ^{2}} d x

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