How do you integrate $\left(\frac{{e}^{x}}{x}\right)dx$ ?

Ashley Bell
2021-12-13
Answered

How do you integrate $\left(\frac{{e}^{x}}{x}\right)dx$ ?

You can still ask an expert for help

Laura Worden

Answered 2021-12-14
Author has **45** answers

This is sometimes called the exponential integral:

$\int \frac{{e}^{x}}{x}dx=Ei\left(x\right)+C$

But the method I'd use (since I'm not familiar with the integral) is the Maclaurin series for$e}^{x$

$e}^{x}=1+x+\frac{{x}^{2}}{2!}+\frac{{x}^{3}}{3!}+\dots =\sum _{n=0}^{\mathrm{\infty}}\frac{{x}^{n}}{n!$

Then:

$\frac{{e}^{x}}{x}=\frac{1}{x}+1+\frac{x}{2!}+\frac{{x}^{2}}{3!}+\dots =\frac{1}{x}+\sum _{n=0}^{\mathrm{\infty}}\frac{{x}^{n}}{(n+1)!}$

So the antiderivative will be:

$\int \frac{{e}^{x}}{x}dx=\int (\frac{1}{x}+1+\frac{x}{2!}+\frac{{x}^{2}}{3!}+\dots )dx$

$=\mathrm{ln}\left(\left|x\right|\right)+x+\frac{{x}^{2}}{2\cdot 2!}+\frac{{x}^{3}}{3\cdot 3!}+\dots +C$

$\int \frac{{e}^{x}}{x}dx=\mathrm{ln}\left(\left|x\right|\right)+\sum _{n=1}^{\mathrm{\infty}}\frac{{x}^{n}}{n\cdot n!}+C$

But the method I'd use (since I'm not familiar with the integral) is the Maclaurin series for

Then:

So the antiderivative will be:

Joseph Lewis

Answered 2021-12-15
Author has **43** answers

Answer: ${e}^{x}=\frac{{x}^{0}}{0}+\frac{{x}^{1}}{1}+\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}+\dots$

$=1+x+\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}+\dots$

$\int \frac{{e}^{x}}{x}dx$

$=\int \frac{1}{x}(1+x+\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}+\dots )dx$

$=\int (\frac{1}{x}+1+\frac{x}{2}+\frac{{x}^{2}}{3}+\dots )dx$

$=\mathrm{log}\left|x\right|+x+\frac{{x}^{2}}{{2}^{2}}+\frac{{x}^{3}}{{3}^{3}}$

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I'm dealing with these two integrals:

${I}_{1}={\int}_{-\pi}^{\pi}\frac{\mathrm{cos}(x)\mathrm{cos}(nx)}{(1+e\mathrm{cos}(x){)}^{3}}\mathrm{d}x,\phantom{\rule{1em}{0ex}}{I}_{2}={\int}_{-\pi}^{\pi}\frac{\mathrm{sin}(x)\mathrm{sin}(nx)}{(1+e\mathrm{cos}(x){)}^{3}}\mathrm{d}x$

is there a way to reduce them to a Hypergeometric form or to solve them analytically?

${I}_{1}={\int}_{-\pi}^{\pi}\frac{\mathrm{cos}(x)\mathrm{cos}(nx)}{(1+e\mathrm{cos}(x){)}^{3}}\mathrm{d}x,\phantom{\rule{1em}{0ex}}{I}_{2}={\int}_{-\pi}^{\pi}\frac{\mathrm{sin}(x)\mathrm{sin}(nx)}{(1+e\mathrm{cos}(x){)}^{3}}\mathrm{d}x$

is there a way to reduce them to a Hypergeometric form or to solve them analytically?

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Why is this solution incorrect?

Prove that on the axis of any parabola there is a certain point K which has the property that, if a chord PQ of the parabola be drawn through it, then$\frac{1}{P{K}^{2}}+\frac{1}{Q{K}^{2}}$ is the same for all positions of the chord.

If it would be valid for standard parabola than it would be valid for all parabolas. Thus, proving for${y}^{2}=4ax$ .

let the point K be (c,0)

Equation of line PQ using parametric coordinates:$x=c+r\mathrm{cos}\theta$ (Equation 1), $y=r\mathrm{sin}\theta$ (Equation 2)

From equation 1 and 2:$(x-c)}^{2}+{y}^{2}={r}^{2$ (Equation 3)

Using Equation 3 and${y}^{2}=4ax$ , we get this quadratic in r: ${r}^{2}-4ax-{(x-k)}^{2}=0$ (Equation 4)

Roots of this quadratic$\left({r}_{1}\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}{r}_{2}\right)$ would be the lengths of PK and QK

From Equation 4,${r}_{1}+{r}_{2}=0$ and ${r}_{1}{r}_{2}=-(4ax+{(x-k)}^{2})$

We know,$\frac{1}{P{K}^{2}}+\frac{1}{Q{K}^{2}}=\frac{1}{{r}_{1}^{2}}+\frac{1}{{r}_{2}^{2}}=\frac{{({r}_{1}+{r}_{2})}^{2}-2{r}_{1}{r}_{2}}{{\left({r}_{1}{r}_{2}\right)}^{2}}=\frac{-2}{{r}_{1}{r}_{2}}$

As value of$r}_{1}{r}_{2$ is not constant thus $\frac{1}{P{K}^{2}}+\frac{1}{Q{K}^{2}}$ does not turn out to be constant. Hence, this solution is incorrect.

I've seen the correct solution but I wanted to know why this solution is incorrect?

Prove that on the axis of any parabola there is a certain point K which has the property that, if a chord PQ of the parabola be drawn through it, then

If it would be valid for standard parabola than it would be valid for all parabolas. Thus, proving for

let the point K be (c,0)

Equation of line PQ using parametric coordinates:

From equation 1 and 2:

Using Equation 3 and

Roots of this quadratic

From Equation 4,

We know,

As value of

I've seen the correct solution but I wanted to know why this solution is incorrect?