What is the integral of 7^{4x+1} ?

The integral of 7^{4x+1} is (7^{4x+1})/(4ln(7))+C

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integral of 7^{4x+1}

\int 7^{4 x + 1} d x

\frac{7 ^{4 x + 1}}{4 ln ( 7 )} + C

Solution steps

\int 7^{4 x + 1} d x

Simplify

7^{4 x + 1} : 7^{4 x} \cdot 7

= \int 7^{4 x} \cdot 7 d x

Take the constant out:

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

= 7 \cdot \int 7^{4 x} d x

Simplify

7^{4 x} : 240 1^{x}

= 7 \cdot \int 240 1^{x} d x

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

= 7 \cdot \frac{240 1 ^{x}}{ln ( 2401 )}

Simplify

7 \cdot \frac{240 1 ^{x}}{ln ( 2401 )} : \frac{7 ^{4 x + 1}}{4 ln ( 7 )}

= \frac{7 ^{4 x + 1}}{4 ln ( 7 )}

Add a constant to the solution

= \frac{7 ^{4 x + 1}}{4 ln ( 7 )} + C

What is the integral of 7^{4x+1} ?
The integral of 7^{4x+1} is (7^{4x+1})/(4ln(7))+C