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

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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

Apply u-substitution

= \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}{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

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