What is the integral of (8x-1)/(2^{4x^2-x)} ?

The integral of (8x-1)/(2^{4x^2-x)} is -1/(ln(2)*2^{4x^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 (8x-1)/(2^{4x^2-x)}

\int \frac{8 x - 1}{2 ^{4 x^{2} - x}} d x

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

Solution steps

\int \frac{8 x - 1}{2 ^{4 x^{2} - x}} d x

Apply u-substitution

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

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

Apply u-substitution

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

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

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}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

\int 2 x^{3} ln (6 x) d x

\int sec (t) (7 sec (t) + 5 tan (t)) d t

\int (sin (5 x))^{2} d x

\int 2^{3} d x

\int \frac{( 2 x - 3 )}{x ^{2} + 6 x + 25} d x

What is the integral of (8x-1)/(2^{4x^2-x)} ?
The integral of (8x-1)/(2^{4x^2-x)} is -1/(ln(2)*2^{4x^2-x)}+C