What is the integral of \sqrt[x]{2} ?

The integral of \sqrt[x]{2} is 2^{1/x}x-ln(2)\Ei(ln(2^{1/x}))+C

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integral of \sqrt[x]{2}

2^{\frac{1}{x}} x - ln (2) Ei (ln (2^{\frac{1}{x}})) + C

Solution steps

Apply u-substitution

: \int - \frac{2 ^{u}}{u ^{2}} d u

= \int - \frac{2 ^{u}}{u ^{2}} d u

Take the constant out:

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

= - \int \frac{2 ^{u}}{u ^{2}} d u

Apply Integration By Parts

: - \frac{2 ^{u}}{u} - \int - \frac{ln ( 2 ) \cdot 2 ^{u}}{u} d u

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

\int - \frac{ln ( 2 ) \cdot 2 ^{u}}{u} d u = - ln (2) Ei (ln (2^{u}))

= - (- \frac{2 ^{u}}{u} - (- ln (2) Ei (ln (2^{u}))))

Substitute back

u = \frac{1}{x}

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

Simplify

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

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

Add a constant to the solution

= 2^{\frac{1}{x}} x - ln (2) Ei (ln (2^{\frac{1}{x}})) + C

f (y) = ln (5 x^{2} - x + 1)

x \to \infty lim (2 x^{2} - x)

\int \frac{7 + t + t ^{2}}{t} d t

\int \frac{v}{3 - 0.001 v ^{2}} d v

x \to \infty lim (\frac{( x e ^{2 x} )}{2})

What is the integral of \sqrt[x]{2} ?
The integral of \sqrt[x]{2} is 2^{1/x}x-ln(2)\Ei(ln(2^{1/x}))+C