What is the integral from 0 to x of (e^{t^2}-1)/t ?

The integral from 0 to x of (e^{t^2}-1)/t is diverges

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integral from 0 to x of (e^{t^2}-1)/t

\int_{0}^{x} \frac{e ^{t^{2}} - 1}{t} d t

diverges

Solution steps

\int_{0}^{x} \frac{e ^{t^{2}} - 1}{t} d t

Expand

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

= \int_{0}^{x} \frac{e ^{t^{2}}}{t} - \frac{1}{t} d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{x} \frac{e ^{t^{2}}}{t} d t - \int_{0}^{x} \frac{1}{t} d t

\int_{0}^{x} \frac{e ^{t^{2}}}{t} d t = \frac{1}{2} (Ei (x^{2}) - Ei (0))

\int_{0}^{x} \frac{1}{t} d t =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

n = 1 \sum \infty \frac{n ^{5}}{3 ^{n}}

\int sec (x) (tan (x) + cos (x)) d x

3 y^{^{'}} = \frac{3 x \cdot y + 18 x ^{2} + 2 y ^{2}}{x ^{2}}, y (1) = 3

f (x) = x^{4} (3 - x)^{2}, at x = 1

\frac{d y}{d x} = 200 - \frac{3 y}{100 - 10 x}

What is the integral from 0 to x of (e^{t^2}-1)/t ?
The integral from 0 to x of (e^{t^2}-1)/t is diverges