What is the integral of ((7^x-1))/x ?

The integral of ((7^x-1))/x is \Ei(ln(7^x))-ln|x|+C

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integral of ((7^x-1))/x

integral

\frac{( 7 ^{x} - 1 )}{x}

Ei (ln (7^{x})) - ln ∣ x ∣ + C

Solution steps

\int \frac{7 ^{x} - 1}{x} d x

Expand

\frac{7 ^{x} - 1}{x} : \frac{7 ^{x}}{x} - \frac{1}{x}

= \int \frac{7 ^{x}}{x} - \frac{1}{x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{7 ^{x}}{x} d x - \int \frac{1}{x} d x

\int \frac{7 ^{x}}{x} d x = Ei (ln (7^{x}))

\int \frac{1}{x} d x = ln ∣ x ∣

= Ei (ln (7^{x})) - ln ∣ x ∣

Add a constant to the solution

= Ei (ln (7^{x})) - ln ∣ x ∣ + C

\frac{x}{4 + x}

\frac{1}{( x ^{2} + 2 x + 4 ) ^{\frac{5}{2}}}

\frac{s}{( s ^{2} + a ^{2} )}

\frac{1}{( x + x ( x ) )}

(tan (x))^{2}

What is the integral of ((7^x-1))/x ?
The integral of ((7^x-1))/x is \Ei(ln(7^x))-ln|x|+C