What is the integral of (2^{x+1})/(10^x) ?

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

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integral of (2^{x+1})/(10^x)

\int \frac{2 ^{x + 1}}{1 0 ^{x}} d x

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

Solution steps

\int \frac{2 ^{x + 1}}{1 0 ^{x}} d x

Simplify

\frac{2 ^{x + 1}}{1 0 ^{x}} : \frac{2}{5 ^{x}}

= \int \frac{2}{5 ^{x}} d x

Take the constant out:

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

= 2 \cdot \int \frac{1}{5 ^{x}} d x

Apply u-substitution

: \int \frac{1}{ln ( 5 ) u ^{2}} d u

= 2 \cdot \int \frac{1}{ln ( 5 ) u ^{2}} d u

Take the constant out:

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

= 2 \cdot \frac{1}{ln ( 5 )} \cdot \int \frac{1}{u ^{2}} d u

Apply the Power Rule

: - \frac{1}{u}

= 2 \cdot \frac{1}{ln ( 5 )} (- \frac{1}{u})

Substitute back

u = 5^{x}

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

Simplify

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

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

Add a constant to the solution

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

\int \frac{1 + sin ( x )}{cos ^{2} ( x )} d x

3 ln (4 - x^{2})

y^{^{'}} = \frac{ln ( 2 )}{7} y - 20, y (0) = 200

\frac{w}{w + \frac{d}{w}}

\frac{d}{d x} (4 y - 2 x)

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