What is the integral of (x-5)/(2^{x-1)} ?

The integral of (x-5)/(2^{x-1)} is 2(-x/(ln(2)*2^x)-1/(ln^2(2)*2^x)+5/(ln(2)*2^x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x-5)/(2^{x-1)}

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

2 (- \frac{x}{ln ( 2 ) \cdot 2 ^{x}} - \frac{1}{ln ^{2} ( 2 ) \cdot 2 ^{x}} + \frac{5}{ln ( 2 ) \cdot 2 ^{x}}) + C

Solution steps

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

\frac{x - 5}{2 ^{x - 1}} = \frac{1}{2 ^{- 1}} \frac{x - 5}{2 ^{x}}

= \int \frac{1}{2 ^{- 1}} \cdot \frac{x - 5}{2 ^{x}} d x

Take the constant out:

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

= \frac{1}{2 ^{- 1}} \cdot \int \frac{x - 5}{2 ^{x}} d x

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

= 2 \cdot \int \frac{x - 5}{2 ^{x}} d x

Expand

\frac{x - 5}{2 ^{x}} : \frac{x}{2 ^{x}} - \frac{5}{2 ^{x}}

= 2 \cdot \int \frac{x}{2 ^{x}} - \frac{5}{2 ^{x}} d x

Apply the Sum Rule:

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

= 2 (\int \frac{x}{2 ^{x}} d x - \int \frac{5}{2 ^{x}} d x)

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

\int \frac{5}{2 ^{x}} d x = - \frac{5}{ln ( 2 ) \cdot 2 ^{x}}

= 2 (- \frac{x}{ln ( 2 ) \cdot 2 ^{x}} - \frac{1}{ln ^{2} ( 2 ) \cdot 2 ^{x}} - (- \frac{5}{ln ( 2 ) \cdot 2 ^{x}}))

Simplify

= 2 (- \frac{x}{ln ( 2 ) \cdot 2 ^{x}} - \frac{1}{ln ^{2} ( 2 ) \cdot 2 ^{x}} + \frac{5}{ln ( 2 ) \cdot 2 ^{x}})

Add a constant to the solution

= 2 (- \frac{x}{ln ( 2 ) \cdot 2 ^{x}} - \frac{1}{ln ^{2} ( 2 ) \cdot 2 ^{x}} + \frac{5}{ln ( 2 ) \cdot 2 ^{x}}) + C

\int sec^{n} (d) x d x

\int 10 - 1.5 x^{\frac{1}{2}} d x

\int x^{4} ln (6 x) d x

\int x e^{(- 5 x)} d x

\int (3 - 18 x^{2}) csc^{2} (2 x^{3} - x) d x

What is the integral of (x-5)/(2^{x-1)} ?
The integral of (x-5)/(2^{x-1)} is 2(-x/(ln(2)*2^x)-1/(ln^2(2)*2^x)+5/(ln(2)*2^x))+C