# Solve\[\] with an Integral

**URL:** <https://community.wolfram.com/t/solve-with-an-integral/4903>\
**Category:** Questions\
**Tags:** mathematica, algebra, calculus\
**Created:** [February 20, 2015, 6:45am UTC](https://community.wolfram.com/t/solve-with-an-integral/4903 "2015-02-20T06:45:51Z")\
**Posts on this page:** 2\
**Page:** 1

<div class="post-metadata">

**Author:** ![gregwhammond](https://avatars.discourse-cdn.com/v4/letter/g/96bed5/32.png) [@gregwhammond](https://community.wolfram.com/u/gregwhammond)\
**Post date:** [February 20, 2015, 6:45am UTC](https://community.wolfram.com/t/solve-with-an-integral/4903/1 "2015-02-20T06:45:51Z")

</div>

Hello,

How can I have this solve for ‘c’? A ‘c’ of around 1 will satisfy the equation.

```auto
a =. 
b = 1; 
c =. 
Solve[Integrate[(-(1/(-1 + E)) - E^a/(1 - E))^(1/(b*c)), {a, 0, 1}] == 1/2, c, Reals]

```

---

<div class="post-metadata">

**Author:** ![mthiel](https://yyz1.discourse-cdn.com/wolfram/user_avatar/community.wolfram.com/mthiel/32/42877_2.png) [@mthiel](https://community.wolfram.com/u/mthiel)\
**Post date:** [February 20, 2015, 8:26am UTC](https://community.wolfram.com/t/solve-with-an-integral/4903/2 "2015-02-20T08:26:05Z")

</div>

Hi Greg,

the integral evaluates to:

```
ConditionalExpression[
 2.71828^(0.458675/c) c Hypergeometric2F1[-(1./c), -(1./c), 1. - 1./c, 0.367879] - 
  2.71828^(-0.541325/c) c Hypergeometric2F1[-(1./c), -(1./c), 1. - 1./c, 1.], Re[1/c] > -1.]

```

So suppose that Solve will not be able to find a non-numerical solution to your equation. If you plot the solution you find:

```
Plot[{Evaluate[2.718281828459045`^(0.4586751453870819`/c)
      c Hypergeometric2F1[-(1.`/c), -(1.`/c), 1.` - 1.`/c, 
      0.36787944117144233`] - 
      2.718281828459045`^(-0.541324854612918`/c)
      c Hypergeometric2F1[-(1.`/c), -(1.`/c), 1.` - 1.`/c, 1.`]], 
  0.5}, {c, 0.1, 1.7}]

```

 ![ScreenShot2015-02-20at08.23.33.png](https://global.discourse-cdn.com/wolfram/original/2X/a/a80fa561994c855672d1888fcc6c62066e3bf27f.png)

You can find the solution by

```
FindRoot[Evaluate[
   1.` 2.718281828459045`^(0.4586751453870819`/c)
      c Hypergeometric2F1[-(1.`/c), -(1.`/c), 1.` - 1.`/c, 
      0.36787944117144233`] - 
    1.` 2.718281828459045`^(-0.541324854612918`/c)
      c Hypergeometric2F1[-(1.`/c), -(1.`/c), 1.` - 1.`/c, 1.`]] == 
  0.5, {c, 1.4}]

```

this gives {c → 1.36961}.

Cheers,

Marco