What is the integral of (e^{x^2}-1)/x ?

The integral of (e^{x^2}-1)/x is 1/2 \Ei(x^2)-ln|x|+C

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integral of (e^{x^2}-1)/x

\int \frac{e ^{x^{2}} - 1}{x} d x

\frac{1}{2} Ei (x^{2}) - ln ∣ x ∣ + C

Solution steps

\int \frac{e ^{x^{2}} - 1}{x} d x

Expand

\frac{e ^{x^{2}} - 1}{x} : \frac{e ^{x^{2}}}{x} - \frac{1}{x}

= \int \frac{e ^{x^{2}}}{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{e ^{x^{2}}}{x} d x - \int \frac{1}{x} d x

\int \frac{e ^{x^{2}}}{x} d x = \frac{1}{2} Ei (x^{2})

\int \frac{1}{x} d x = ln ∣ x ∣

= \frac{1}{2} Ei (x^{2}) - ln ∣ x ∣

Add a constant to the solution

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

\frac{d}{d x} (a (2 e^{- x} x - e^{- x} x^{2}))

x \to 1 lim (9 x - 5)

\frac{d}{d x} (x^{3} + x^{2} - 6 x)

h (x) = 2 x (6 x - 18)^{6}

\int \frac{7 - 4 x ^{3} + 6 x ^{6}}{x ^{6}} d x

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