What is the integral of 10^{5x-1} ?

The integral of 10^{5x-1} is (2^{5x-1}*5^{5x-2})/(ln(10))+C

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integral of 10^{5x-1}

\int 1 0^{5 x - 1} d x

\frac{2 ^{5 x - 1} \cdot 5 ^{5 x - 2}}{ln ( 10 )} + C

Solution steps

\int 1 0^{5 x - 1} d x

1 0^{5 x - 1} = 1 0^{- 1} \cdot 1 0^{5 x}

= \int 1 0^{- 1} \cdot 1 0^{5 x} d x

Take the constant out:

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

= 1 0^{- 1} \cdot \int 1 0^{5 x} d x

Apply exponent rule:

a^{- 1} = \frac{1}{a}

= \frac{1}{10} \cdot \int 1 0^{5 x} d x

Simplify

1 0^{5 x} : 10000 0^{x}

= \frac{1}{10} \cdot \int 10000 0^{x} d x

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

= \frac{1}{10} \cdot \frac{10000 0 ^{x}}{ln ( 100000 )}

Simplify

\frac{1}{10} \cdot \frac{10000 0 ^{x}}{ln ( 100000 )} : \frac{2 ^{5 x - 1} \cdot 5 ^{5 x - 2}}{ln ( 10 )}

= \frac{2 ^{5 x - 1} \cdot 5 ^{5 x - 2}}{ln ( 10 )}

Add a constant to the solution

= \frac{2 ^{5 x - 1} \cdot 5 ^{5 x - 2}}{ln ( 10 )} + C

\int_{1}^{\infty} \frac{1}{11 x + 7} d x

\frac{\partial}{\partial x} (x^{8} + 2 x y^{7})

t^{4} - 8 t^{3} + 1 0^{2}

2 cos (5 t)

y = e^{x}, y = e^{- x}, 0, 4

What is the integral of 10^{5x-1} ?
The integral of 10^{5x-1} is (2^{5x-1}*5^{5x-2})/(ln(10))+C