What is the integral of (2^{(18x)})/4 ?

The integral of (2^{(18x)})/4 is (2^{18x-3})/(9ln(2))+C

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integral of (2^{(18x)})/4

\int \frac{2 ^{(18 x)}}{4} d x

\frac{2 ^{18 x - 3}}{9 ln ( 2 )} + C

Solution steps

\int \frac{2 ^{18 x}}{4} d x

Take the constant out:

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

= \frac{1}{4} \cdot \int 2^{18 x} d x

Apply Integration By Parts

: 26214 4^{x} x - \int 9 ln (2) \cdot 2^{18 x + 1} x d x

= \frac{1}{4} (26214 4^{x} x - \int 9 ln (2) \cdot 2^{18 x + 1} x d x)

\int 9 ln (2) \cdot 2^{18 x + 1} x d x = 26214 4^{x} x - \frac{2 ^{18 x - 1}}{9 ln ( 2 )}

= \frac{1}{4} (26214 4^{x} x - (26214 4^{x} x - \frac{2 ^{18 x - 1}}{9 ln ( 2 )}))

Simplify

\frac{1}{4} (26214 4^{x} x - (26214 4^{x} x - \frac{2 ^{18 x - 1}}{9 ln ( 2 )})) : \frac{2 ^{18 x - 3}}{9 ln ( 2 )}

= \frac{2 ^{18 x - 3}}{9 ln ( 2 )}

Add a constant to the solution

= \frac{2 ^{18 x - 3}}{9 ln ( 2 )} + C

\int 2 x x d x

\frac{d}{d x} ((9 x^{3} + 3 x^{2})^{4})

- y^{^{'}} - 5 y = - t

\frac{d y}{d x} + 10 (tan (10 x)) y = cos (10 x)

\frac{d y}{d x} = 4 x^{3} y

What is the integral of (2^{(18x)})/4 ?
The integral of (2^{(18x)})/4 is (2^{18x-3})/(9ln(2))+C