What is the integral of 4^{4x-1} ?

The integral of 4^{4x-1} is 4^{4x-1}x-(2^{8x-2}x-(2^{8x-5})/(ln(2)))+C

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integral of 4^{4x-1}

\int 4^{4 x - 1} d x

4^{4 x - 1} x - (2^{8 x - 2} x - \frac{2 ^{8 x - 5}}{ln ( 2 )}) + C

Solution steps

\int 4^{4 x - 1} d x

Apply Integration By Parts

: 4^{4 x - 1} x - \int ln (2) \cdot 2^{8 x + 1} x d x

= 4^{4 x - 1} x - \int ln (2) \cdot 2^{8 x + 1} x d x

\int ln (2) \cdot 2^{8 x + 1} x d x = 2^{8 x - 2} x - \frac{2 ^{8 x - 5}}{ln ( 2 )}

= 4^{4 x - 1} x - (2^{8 x - 2} x - \frac{2 ^{8 x - 5}}{ln ( 2 )})

Add a constant to the solution

= 4^{4 x - 1} x - (2^{8 x - 2} x - \frac{2 ^{8 x - 5}}{ln ( 2 )}) + C

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What is the integral of 4^{4x-1} ?
The integral of 4^{4x-1} is 4^{4x-1}x-(2^{8x-2}x-(2^{8x-5})/(ln(2)))+C