How can I integrate e^[a*sqrt(b^2+x^2)] * e^c*x?

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In summary, Integrate e^[a*sqrt(b^2+x^2)] * e^c*x is difficult because the exponent of the first e is not simply x.
  • #1
phys_student1
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Integrate e^[a*sqrt(b^2+x^2)] * e^c*x

I can integrate the first part alone, i.e. ∫ea*√(b2+x2) and the result is [√(b2+x2)/a - 1/a2] ea*√(b2+x2).

For the full integral, i.e. ∫ea*√(b2+x2) * ec*x I tried using integration by parts but it does not work since the exponent of the first e is not simply x.

the second exponent was originally a cos but that is still much harder.

Any ideas?
 
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  • #2
Strange, I changed the title but it looks like it did not.
 
  • #3
It also helps to know with respect to which variable the integration is to be done.
 
  • #4
SteamKing said:
It also helps to know with respect to which variable the integration is to be done.

wrt x.
 
  • #5
Let's start by taking all of the constants out of the integral:

e^(a + c)INT[e^sqrt(b^2+x^2)*x,dx]

Now we need to expand out the x to "sqrt(b^2 + x^2)(x/sqrt(b^2 + x^2))". The reason for this will become apparent.

e^(a + c)INT[e^sqrt(b^2+x^2)*sqrt(b^2 + x^2)(x/sqrt(b^2 + x^2)),dx]

Now we use U substitution where:

u = sqrt(b^2 + x^2)
du/dx = x/sqrt(b^2 + x^2)

So:

e^(a + c)INT[ue^u,du]

Now we integrate by parts:

e^(a + c)(ue^u + INT[e^u,du])

And solve:

e^(a + c)(ue^u + e^u)

Simplify:

e^(a + c)((u + 1)e^u)
(u + 1)e^(u + a + c)

Un-substitute the u:

(sqrt(b^2 + x^2) + 1)e^(sqrt(b^2 + x^2) + a + c)

There may be simplifying to be done, but you should be able to manage that. Please check over my work to see if there is anything that looks wrong before trusting me, sometimes I confuse myself when I need to format things like this with my keyboard.
 
  • #6
LastTimelord said:
Let's start by taking all of the constants out of the integral:

e^(a + c)INT[e^sqrt(b^2+x^2)*x,dx]

Now we need to expand out the x to "sqrt(b^2 + x^2)(x/sqrt(b^2 + x^2))". The reason for this will become apparent.

e^(a + c)INT[e^sqrt(b^2+x^2)*sqrt(b^2 + x^2)(x/sqrt(b^2 + x^2)),dx]

Now we use U substitution where:

u = sqrt(b^2 + x^2)
du/dx = x/sqrt(b^2 + x^2)

So:

e^(a + c)INT[ue^u,du]

Now we integrate by parts:

e^(a + c)(ue^u + INT[e^u,du])

And solve:

e^(a + c)(ue^u + e^u)

Simplify:

e^(a + c)((u + 1)e^u)
(u + 1)e^(u + a + c)

Un-substitute the u:

(sqrt(b^2 + x^2) + 1)e^(sqrt(b^2 + x^2) + a + c)

There may be simplifying to be done, but you should be able to manage that. Please check over my work to see if there is anything that looks wrong before trusting me, sometimes I confuse myself when I need to format things like this with my keyboard.

You cannot take e^c out. C is multiplied by the square root, not added. Same for e^a.
 
  • #7
You can take out any coefficients. If it were added, it would need to have an x multiplied by it, but if the equation is a monomial, you can take any constant coefficients out of the integral, without affecting the answer.
 
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  • #8
see the difference:

2yp0sc2.gif
 
  • #9
It's e^c*x, so it's logical to interpret it as "(e^c)*x", instead of what you seem to be doing: "e^(c*x), but I do now see what I did wrong: For some reason I forgot my exponent rules and thought "a^(bc) = (a^b)(a^c), when indeed it does not. Sorry, I may have been previously misunderstanding your point.

This is the correct solution:

INT[e^(a*sqrt(b^2+x^2)) * e^c*x,dx]

taking out constants, this time correctly:

(e^c)INT[e^(a*sqrt(b^2+x^2)) *x,dx]

Expanding x:

(e^c)INT[e^(a*sqrt(b^2+x^2))*(sqrt(b^2 + x^2)/a)*(ax/sqrt(b^2 + a^2)),dx]

U substitution:

u = a*sqrt(b^2+x^2)
u/a^2 = sqrt(b^2+x^2)/a
du/dx = ax/sqrt(b^2 + a^2)
du = ax/sqrt(b^2 + a^2)dx

(e^c)INT[e^(u)*(u/a^2),du]

Taking out constants:

(e^c)/(a^2)INT[u*e^(u),du]

Integration by parts:

(e^c)/(a^2)(ue^u + INT[e^u,du])

And evaluate the integral:

(e^c)/(a^2)(ue^u + e^u) + C

Un-substitute:

(e^c)/(a^2)((a*sqrt(b^2+x^2))e^(a*sqrt(b^2+x^2)) + e^(a*sqrt(b^2+x^2))) + C

Simplify:

(e^c)/(a^2)(a*sqrt(b^2+x^2) + 1)e^(a*sqrt(b^2+x^2)) + C

There may be more simplifying to do, but I won't get into that.
 
  • #10
Thank you but I want e to the power of c*x.

But anyway thank you indeed.
 
  • #11
Hi !
Integral of exp[a*sqrt(b^2+x^2)]*exp(c*x)*dx
this integral cannot be expressed in terms of a finite number of standard functions, but in terms of tough series of hyperbolic functions. See attachment.
The formal result will be so ardeous that I doubt that it could be useful for further calculus in practice.
So, I suggest to use numerical methods instead of analytical.
 

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  • #12
JJacquelin said:
Hi !
Integral of exp[a*sqrt(b^2+x^2)]*exp(c*x)*dx
this integral cannot be expressed in terms of a finite number of standard functions, but in terms of tough series of hyperbolic functions. See attachment.
The formal result will be so ardeous that I doubt that it could be useful for further calculus in practice.
So, I suggest to use numerical methods instead of analytical.

Thanks indeed!

Yes I tried a lot but have not find any good result and you confirmed this. Thanks for your effort, I really appreciate your time! Thanks again.
 

1. What is the general formula for integrating e^[a*sqrt(b^2+x^2)]?

The general formula for integrating e^[a*sqrt(b^2+x^2)] is ∫e^[a*sqrt(b^2+x^2)]dx = (x/a)*e^[a*sqrt(b^2+x^2)] + C.

2. How do you solve an integral with e^[a*sqrt(b^2+x^2)] in the integrand?

To solve the integral, use the substitution method by letting u = sqrt(b^2+x^2) and finding the derivative du/dx. Then, replace sqrt(b^2+x^2) with u and dx with du in the integral. This will result in an integral with only e^u, which can be easily integrated.

3. Can the integral e^[a*sqrt(b^2+x^2)] be solved analytically?

Yes, the integral can be solved analytically using the substitution method described above.

4. Are there any special cases when integrating e^[a*sqrt(b^2+x^2)]?

Yes, when a=0, the integral becomes ∫e^[0*sqrt(b^2+x^2)]dx = ∫dx = x + C. When b=0, the integral becomes ∫e^[a*sqrt(x^2)]dx = ∫e^[ax]dx, which can be solved using integration by parts.

5. Can the integral e^[a*sqrt(b^2+x^2)] be used to solve real-world problems?

Yes, this integral can be used in various fields of science and engineering, such as in calculating electric potential due to point charges, solving problems in quantum mechanics, and finding the area under a Gaussian curve.

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