What is the integral from 1 to 2 of (2^x)/x ?

The integral from 1 to 2 of (2^x)/x is \Ei(2ln(2))-\Ei(ln(2))

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integral from 1 to 2 of (2^x)/x

\int_{1}^{2} \frac{2 ^{x}}{x} d x

Ei (2 ln (2)) - Ei (ln (2))

Solution steps

\int_{1}^{2} \frac{2 ^{x}}{x} d x

Apply u-substitution

: \int_{2}^{4} \frac{1}{ln ( u )} d u

= \int_{2}^{4} \frac{1}{ln ( u )} d u

Apply u-substitution

: \int_{l n (2)}^{2 l n (2)} \frac{e ^{v}}{v} d v

= \int_{l n (2)}^{2 l n (2)} \frac{e ^{v}}{v} d v

Use the nonelementary integral:

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

= [Ei (v)]_{l n (2)}^{2 l n (2)}

Compute the boundaries:

Ei (2 ln (2)) - Ei (ln (2))

= Ei (2 ln (2)) - Ei (ln (2))

\int_{- 2 x}^{x} \frac{1}{csc ^{5} ( t )} d t

\int_{0}^{2} 5 x^{2} - 8 x + 4 d x

\int_{1}^{5} \frac{2}{t ^{3}} d t

\int_{- 1}^{1} 1 + x + x^{2} + x^{3} d x

\int_{- 1}^{1} (1 - x) 1 - x^{2} d x

What is the integral from 1 to 2 of (2^x)/x ?
The integral from 1 to 2 of (2^x)/x is \Ei(2ln(2))-\Ei(ln(2))