# Can't solve this difficult definite integral for the life of me

• dumbperson
In summary: Well, you'll figure it out ;-)How do you come up with a substitution like this?One possible way is to look for a trigonometric substitution. Since we have a fraction with a square in the denominator, it might work to try something like x=arctan(u) or x=arcsin(u).Another way is to try a substitution that transforms the given integral into a simpler form. In this case, we have a fraction with a logarithm in the numerator and a polynomial in the denominator. So we could try to transform it into a fraction with just a polynomial in both the numerator and denominator. The substitution u = (x+1)^2 might be a good candidate for this.Finally,
dumbperson

## Homework Statement

Definite integral:

$$\int \frac{\ln(x+1)}{x^2+1} dx$$ from x=0 to x=1 (sorry I don't know how to do integral boundaries with tex)

## The Attempt at a Solution

I just am clueless on how to do this, I'm almost 100% sure you are not supposed to find the anti-derivative of the integral and then calculate the definite integral, but need a clever trick
to calculate the definite integral.

So the standard tricks like u-substition and partial integration do not work.

Any tips?
Thanks!

It isn't an elementary integral. Maple gives an answer in terms of dilog functions and a Catalan number.

I would be inclined to use a numerical integration.

It has an algebraic solution and must be solved algebraically(so no numerical methods). I already said I thought it was not an elementary integral , and said that this one has to be solved in a clever way. Have been stuck on this one for almost a week. The answer is (pi/8)ln(2)

HallsofIvy said:
I would be inclined to use a numerical integration.
See my reply, sorry forgot to quote.

dumbperson said:

## Homework Statement

Definite integral:$$\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} dx$$Use _{0} and ^{1} for the limits of integration.

## The Attempt at a Solution

I just am clueless on how to do this, I'm almost 100% sure you are not supposed to find the anti-derivative of the integral and then calculate the definite integral, but need a clever trick
to calculate the definite integral.

So the standard tricks like u-substitution and partial integration do not work.

Any tips?
Thanks!

Try a u-substitution that takes x=0 to u=1 and x=1 to u=0 .

Possibilities:

u = 1-t (but I think this won't work.)

[STRIKE]u = 1/(x+1) - 1 → u+1 = 1/(x+1)[/STRIKE]

[STRIKE]u = 1/(x-1) + 1 → u-1 = 1/(x-1)[/STRIKE]

(Those last two aren't as advertized. DUH !)

Try $\displaystyle \ \ u=\frac{1-x}{1+x}\ .$

Last edited:
SammyS said:
Try a u-substitution that takes x=0 to u=1 and x=1 to u=0 .

Possibilities:

u = 1-t (but I think this won't work.)

[STRIKE]u = 1/(x+1) - 1 → u+1 = 1/(x+1)[/STRIKE]

[STRIKE]u = 1/(x-1) + 1 → u-1 = 1/(x-1)[/STRIKE]

(Those last two aren't as advertized. DUH !)

Try $\displaystyle \ \ u=\frac{1-x}{1+x}\ .$
I got stuck using that substitution, it just got very messy

If you integrate by parts in the obvious way, the first term gives, intriguingly, exactly twice the desired answer. The second term is -∫atan(x)/(1+x).dx.
By a series of substitutions (x = tan(θ), θ→θ/2, θ→θ-π/4; amounting, I think, to x = tan(θ/2-π/8)) I turned that into some multiple of $\int_{\pi/4}^{3 \pi/4}\frac{\left(\theta-\pi/4\right)}{sin(\theta)}.d\theta$. Splitting the range around π/2 and combining the θ term with its π-θ partner turned that into integrating cosec. Haven't checked it produces the right answer.

I notice you have 1/(x^2+1). Try a u-substitution where u= 1/(x^2+1).

dumbperson said:
I got stuck using that substitution, it just got very messy.
For those of you helping out, as LCKurtz noted in post #2, the indefinite integral cannot be expressed in terms of elementary functions. However, the definite integral can be evaluated using a "trick".

The result of the substitution $\displaystyle \ \ u=\frac{1-x}{1+x}\,,\$ is fairly clean.

$\displaystyle \ \ u=\frac{1-x}{1+x}=\frac{2}{1+x}-1$

So that $\displaystyle \ \ u+1=\frac{2}{1+x}\ .\ \$ This substitution behaves quite nicely.

This gives $\displaystyle \ \ x+1=\frac{2}{1+u}\,,\ \$ so finding dx is easy.

Also note that $\displaystyle \ \ 1+x^2=1+\left(\frac{1-u}{1+u}\right)^2=2\frac{1+u^2}{(1+u)^2}\ .$ Therefore, lots of things cancel out.

SammyS said:
For those of you helping out, as LCKurtz noted in post #2, the indefinite integral cannot be expressed in terms of elementary functions. However, the definite integral can be evaluated using a "trick".

The result of the substitution $\displaystyle \ \ u=\frac{1-x}{1+x}\,,\$ is fairly clean.

$\displaystyle \ \ u=\frac{1-x}{1+x}=\frac{2}{1+x}-1$

So that $\displaystyle \ \ u+1=\frac{2}{1+x}\ .\ \$ This substitution behaves quite nicely.

This gives $\displaystyle \ \ x+1=\frac{2}{1+u}\,,\ \$ so finding dx is easy.

Also note that $\displaystyle \ \ 1+x^2=1+\left(\frac{1-u}{1+u}\right)^2=2\frac{1+u^2}{(1+u)^2}\ .$ Therefore, lots of things cancel out.

Ok, I shall try again.

$$dx = \frac{-2}{(1+u)^2} du$$

So the integral becomes

$$\int_{1}^{0} \frac{\ln(\frac{2}{1+u})}{\frac{2(1+u^2}{(1+u)^2}} \frac{-2}{(1+u)²} du = -\int_{1}^{0} \frac{\ln(\frac{2}{1+u})}{1+u^2} du = \int_{0}^{1} \frac{\ln(\frac{2}{1+u})}{1+u^2} du$$

I don't see how I can get the definite integral with this substitution :-(

Try to simplify ln(2/(1+u)), and compare it with the integral you started with.

dumbperson said:
Ok, I shall try again.

$$dx = \frac{-2}{(1+u)^2} du$$

So the integral becomes

$$\int_{1}^{0} \frac{\ln(\frac{2}{1+u})}{\frac{2(1+u^2}{(1+u)^2}} \frac{-2}{(1+u)²} du = -\int_{1}^{0} \frac{\ln(\frac{2}{1+u})}{1+u^2} du = \int_{0}^{1} \frac{\ln(\frac{2}{1+u})}{1+u^2} du$$

I don't see how I can get the definite integral with this substitution :-(

You are practically there. Use the rules of logs to break the integral into two parts. Notice something about one of them.

Dick said:
You are practically there. Use the rules of logs to break the integral into two parts. Notice something about one of them.

Oh Wow, got it now. How does one come up with this substitution?

SammyS said:
For those of you helping out, as LCKurtz noted in post #2, the indefinite integral cannot be expressed in terms of elementary functions. However, the definite integral can be evaluated using a "trick".

The result of the substitution $\displaystyle \ \ u=\frac{1-x}{1+x}\,,\$ is fairly clean.

$\displaystyle \ \ u=\frac{1-x}{1+x}=\frac{2}{1+x}-1$

So that $\displaystyle \ \ u+1=\frac{2}{1+x}\ .\ \$ This substitution behaves quite nicely.

This gives $\displaystyle \ \ x+1=\frac{2}{1+u}\,,\ \$ so finding dx is easy.

Also note that $\displaystyle \ \ 1+x^2=1+\left(\frac{1-u}{1+u}\right)^2=2\frac{1+u^2}{(1+u)^2}\ .$ Therefore, lots of things cancel out.

Thanks ! I got it now, how do you come up with a substitution like this?

dumbperson said:
how do you come up with a substitution like this?
A frequent trick with nasty definite integrals is to find a way to fold the range so that the nastiness in one part cancels the nastiness in the rest. Or, what is really the same thing, reverse the whole range and reform it so that the original integral pops out as only part of the expression, and with a factor other than 1 (in this case, -1).
Since the ln(x+1) was the nasty part, that's the part to study for working out the point around which to pivot the range. To make it easier, you could first write x-1 wherever there's an x, so we get ln(x), and the range becomes 1 to 2. A natural way to pivot the range given ln(x) is to map x to 1/x, so ln(x) becomes -ln(x).
The present problem generalises to ##\int_a^b\frac{ln(x)dx}{x^2+Bx+ab}##

haruspex said:
A frequent trick with nasty definite integrals is to find a way to fold the range so that the nastiness in one part cancels the nastiness in the rest. Or, what is really the same thing, reverse the whole range and reform it so that the original integral pops out as only part of the expression, and with a factor other than 1 (in this case, -1).
Since the ln(x+1) was the nasty part, that's the part to study for working out the point around which to pivot the range. To make it easier, you could first write x-1 wherever there's an x, so we get ln(x), and the range becomes 1 to 2. A natural way to pivot the range given ln(x) is to map x to 1/x, so ln(x) becomes -ln(x).
The present problem generalises to ##\int_a^b\frac{ln(x)dx}{x^2+Bx+ab}##

Okay, thanks!

But is there a way to see beforehand what would maybe work and what wouldn't ? or do you just have to try?

## 1. How do I approach solving a difficult definite integral?

There are several strategies you can use to approach a difficult definite integral. These include substitution, integration by parts, partial fractions, and trigonometric identities. It is important to carefully analyze the integral and choose the most appropriate method for solving it.

## 2. What should I do if I get stuck while trying to solve a difficult definite integral?

If you get stuck while trying to solve a difficult definite integral, take a break and come back to it later with a fresh mind. You can also try approaching the integral from a different angle or seeking help from a peer or teacher.

## 3. Is it possible that a definite integral is unsolvable?

Yes, there are some definite integrals that are considered unsolvable. These are known as "special functions" and require advanced mathematical techniques to solve.

## 4. Can technology be used to solve difficult definite integrals?

Yes, technology such as calculators and computer software can be used to solve difficult definite integrals. However, it is important to understand the steps involved in solving the integral manually in order to fully understand the concept.

## 5. Are there any tips or tricks for solving difficult definite integrals?

Yes, some tips and tricks for solving difficult definite integrals include breaking the integral into smaller parts, using symmetry to simplify the integral, and choosing the most appropriate method for solving based on the form of the integral.

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