What is the integral of (e^{x^n})/(x^{1-n)} ?

The integral of (e^{x^n})/(x^{1-n)} is 1/n e^{x^n}+C

Popular >

integral of (e^{x^n})/(x^{1-n)}

\int \frac{e ^{x^{n}}}{x ^{1 - n}} d x

\frac{1}{n} e^{x^{n}} + C

Solution steps

\int \frac{e ^{x^{n}}}{x ^{1 - n}} d x

Apply u-substitution

: \int \frac{e ^{u}}{n} d u

= \int \frac{e ^{u}}{n} d u

Take the constant out:

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

= \frac{1}{n} \cdot \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= \frac{1}{n} e^{u}

Substitute back

u = x^{n}

= \frac{1}{n} e^{x^{n}}

Add a constant to the solution

= \frac{1}{n} e^{x^{n}} + C

h \to 0 lim (\frac{5 e ^{x} - 5 e ^{x + h}}{3 h})

\frac{e ^{- s}}{( s ^{2} + 8 )}

\frac{d}{d x} (\frac{3 x - 1}{2 x + 1})

y^{^{'}} + 4 x^{- 1} y = x^{2}, y (1) = - 2

n = 1 \sum \infty \frac{5 ^{n}}{n ^{5}}

What is the integral of (e^{x^n})/(x^{1-n)} ?
The integral of (e^{x^n})/(x^{1-n)} is 1/n e^{x^n}+C