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

\int_{0}^{1} \frac{1}{( x - 1 ) ^{2}} d x

\frac{1 + x}{1 - x}

\int_{0}^{3} π (2 x)^{2} d x

\int \frac{1}{( 1 - y ^{2} ) ^{\frac{1}{2}}} d y

\int \frac{x + 1}{x ^{2} + 2 x + 2} 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