# Evaluate The Integrals

#### GreenGoblin

##### Member
Please help me to evaluate the following integrals:

1) $\int\frac{x^{4}+1}{x^{2}+1}dx$

I recognise the form of $x^{2}+1$ in the denominator corresponds to an inverse tangent derivative. But how would I deal with the numerator in this respect?

2) $\int\frac{1}{x^{2}+x-6}dx$

I believe this involves completing the square, I made the first step of doing this, rearranging to $\int\frac{1}{(x + \frac{1}{2})^{2}-\frac{25}{4}}dx$, but I am not entirely sure of the exact integration formula corresponding to this. I observe 25 and 4 are clearly both squares too so I assume this problem set this up intentionally, and this bears a relevance for the remainder of the problem.

Gracias,
GreenGoblin

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#### chisigma

##### Well-known member
Please help me to evaluate the following integrals:

1) $\int\frac{x^{4}+1}{x^{2}+1}dx$

I recognise the form of $x^{2}+1$ in the denominator corresponds to an inverse tangent derivative. But how would I deal with the numerator in this respect?

Gracias,
GreenGoblin
$\displaystyle \frac{1+x^{4}}{1+x^{2}}= x^{2}-1+\frac{2}{1+x^{2}}$

Kind regards

$\chi$ $\sigma$

#### CaptainBlack

##### Well-known member
2) $\int\frac{1}{x^{2}+x-6}dx$
Partial fractions:

$\int\frac{1}{x^{2}+x-6}dx ={\Large{\int}} \left[-\frac{1}{5(x+3)}+\frac{1}{5(x-2)}\right]\; dx$

CB

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#### soroban

##### Well-known member
Hello, GreenGoblin!

$2)\;\int\dfrac{1}{x^{2}+x-6}dx$

I believe this involves completing the square,
I made the first step of doing this, rearranging to $\int\frac{1}{(x + \frac{1}{2})^{2}-\frac{25}{4}}dx$
but I am not entirely sure of the exact integration formula corresponding to this.

You have the inverse secant form.

Let $x+\frac{1}{2} \:=\:\frac{5}{2}\sec\theta \quad\Rightarrow\quad dx \:=\:\frac{5}{2}\sec\theta\tan\theta\,d\theta$
. . . And the denominator becomes: $\frac{25}{4}\tan^2\theta$

$\displaystyle\text{Substitute: }\:\int\dfrac{\frac{5}{2}\sec\theta\tan\theta\,d\theta}{\frac{25}{4}\tan^2\theta} \;=\;\tfrac{2}{5}\int\frac{d\theta}{\sin\theta}$

. . . . . . . . $\displaystyle=\;\tfrac{2}{5}\int \csc\theta\,d\theta \;=\; \tfrac{2}{5}\ln|\csc\theta - \cot\theta| + C$

Back-substitute: .$\sec\theta \:=\:\dfrac{x+\frac{1}{2}}{\frac{5}{2}} \:=\:\dfrac{hyp}{adj}$

$\theta\text{ is in a right triangle with: }\,adj = \frac{5}{2},\;hyp = x + \frac{1}{2}$

Then: .$opp \,=\,\sqrt{(x+\frac{1}{2})^2 - (\frac{5}{2})^2} \:=\:\sqrt{x^2+x-6}$

Hence: .$\csc\theta \:=\:\dfrac{hyp}{opp} \:=\:\dfrac{x+\frac{1}{2}}{\sqrt{x^2+x-6}},\;\;\cot\theta \:=\:\dfrac{adj}{opp} \:=\:\dfrac{\frac{5}{2}}{\sqrt{x^2+x-6}}$

$\text{The solution is: }\:\dfrac{2}{5}\ln\left|\dfrac{x+\frac{1}{2}}{\sqrt{x^2+x-6}} - \dfrac{\frac{5}{2}}{\sqrt{x^2+x-6}}\right| + C \;=\; \dfrac{2}{5}\ln\left|\dfrac{x-2}{\sqrt{x^2+x-6}}\right| + C$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Why is it printing those things in red?

Evidently the program thinks I've made an error
. . and it's pointing it out.
Well, I've made no error ... I've re-type it.
. . I've even re-type the entire statement.
. . It still tells me that it's wrong.
I've never had any problems with Latex before.
. . What's going on?

#### Sherlock

##### Member

$\text{The solution is: }\:\dfrac{2}{5}\ln\left|\dfrac{x+\frac{1}{2}}{ \sqrt{x^2+x-6}} - \dfrac{\frac{5}{2}}{\sqrt{x^2+x-6}}\right| + C \;=\; \dfrac{2}{5}\ln\left|\dfrac{x-2}{\sqrt{x^2+x-6}}\right| + C$
Putting a space before the red term usually takes it away (like it did above).

#### HallsofIvy

##### Well-known member
MHB Math Helper
Sherlock, I am sure, means after \sqrt. The LaTeX does not recognise \sqrtx. You need \sqrt x.

#### Sherlock

##### Member
Soroban earlier (now he changed it but put the space in the wrong place) had...

\text{The solution is: }\:\dfrac{2}{5}\ln\left|\dfrac{x+\frac{1}{2}}{\sqrt{x^2+x-6}} - \dfrac{\frac{5}{2}}{\sqrt{x^2+x-6}}\right| + C \;=\; \dfrac{2}{5}\ln\left|\dfrac{x-2}{\sqrt{x^2+x-6}}\right| + C

... which is absolutely fine, but it didn't render well somehow:

$\text{The solution is: }\:\dfrac{2}{5}\ln\left|\dfrac{x+\frac{1}{2}}{\sqrt{x^2+x-6}} - \dfrac{\frac{5}{2}}{\sqrt{x^2+x-6}}\right| + C \;=\; \dfrac{2}{5}\ln\left|\dfrac{x-2}{\sqrt{x^2+x-6}}\right| + C$

All I did was to put space in here
{\sqrt{x^2+x-6}} and turned it into { \sqrt{x^2+x-6}}

$\text{The solution is: }\:\dfrac{2}{5}\ln\left|\dfrac{x+\frac{1}{2}}{ \sqrt{x^2+x-6}} - \dfrac{\frac{5}{2}}{\sqrt{x^2+x-6}}\right| + C \;=\; \dfrac{2}{5}\ln\left|\dfrac{x-2}{\sqrt{x^2+x-6}}\right| + C$

It did that to me many times but I figured spacing does the trick.

#### GreenGoblin

##### Member
Hello,

$\int \frac{1}{(\frac{b}{a})^{2}-x^{2}}dx = (\frac{a}{b})tanh^{-1}(\frac{a}{b})$

Is it not? Is this something I should use here? If I take a -1 outside of the integral, I have this form. Which is the better approach, this or those others suggested? I see there is more than one way to approach this problem.

As for the partial fractions, can someone explain to me how that rearranges to that form please?

#### Also sprach Zarathustra

##### Member
Hello,

$\int \frac{1}{(\frac{b}{a})^{2}-x^{2}}dx = (\frac{a}{b})tanh^{-1}(\frac{a}{b})$

Is it not? Is this something I should use here? If I take a -1 outside of the integral, I have this form. Which is the better approach, this or those others suggested? I see there is more than one way to approach this problem.

As for the partial fractions, can someone explain to me how that rearranges to that form please?

Return to your algebra books from high-school!

#### GreenGoblin

##### Member
Return to your algebra books from high-school!
This is an unneccesary comment, also makes a couple of unfounded assumptions. Regardless you need not be insulting, I asked a genuine question.

#### Krizalid

##### Active member
I think he meant it about your last line, so he's right and probably someone woulda told you something related as using google or check your books.

#### CaptainBlack

##### Well-known member
As for the partial fractions, can someone explain to me how that rearranges to that form please?
$\frac{1}{x^2+x-6}=\frac{1}{(x-2)(x+3)}=\frac{A}{x-2}+\frac{B}{x+3}$

Now it is just algebra to establish what $$A$$ and $$B$$ need to be.

CB

#### Random Variable

##### Well-known member
MHB Math Helper
Return to your algebra books from high-school!
Your response was totally unnecessary.

I didn't learn about partial fractions until my second semester of calculus.

#### Also sprach Zarathustra

##### Member
Your response was totally unnecessary.

I didn't learn about partial fractions until my second semester of calculus.

What do you want?

#### Random Variable

##### Well-known member
MHB Math Helper
What do you want?
GreenGolbin clearly felt insulted by your response. Was your intention to insult him?

And partial fractions is not something that is usually covered in an high school algebra class. It's probably covered in a later section in his calculus book.

#### Sherlock

##### Member
You don't have to know the method of partial fractions, to be honest -- as:

$\displaystyle \frac{1}{x^2+x-6} = \frac{1}{(x+3)(x-2)} = \frac{(x+3)-(x-2)}{5(x+3)(x-2)} = \frac{1}{5(x-2)}-\frac{1}{5(x+3)}$

Moreover, you don't even have to use that -- you can let $t = \frac{x+3}{x-2}$ or $t = \frac{x-2}{x+3}$.

By the way, ASZ might have been genuinely suggesting to search this in a book.
(Sometimes there is a language barrier that maligns a perfectly innocent tone).

#### Random Variable

##### Well-known member
MHB Math Helper
You don't have to know the method of partial fractions, to be honest -- as:

$\displaystyle \frac{1}{x^2+x-6} = \frac{1}{(x+3)(x-2)} = \frac{(x+3)-(x-2)}{5(x+3)(x-2)} = \frac{1}{5(x-2)}-\frac{1}{5(x+3)}$

Moreover, you don't even have to use that -- you can let $t = \frac{x+3}{x-2}$ or $t = \frac{x-2}{x+3}$.

By the way, ASZ might have been genuinely suggesting to search this in a book.
(Sometimes there is a language barrier that maligns a perfectly innocent tone).
Perhaps. I just don't like anyone to feel that they're being mocked or ridiculed unless they truly deserve it.

Interesting substitutions, BTW.

#### GreenGoblin

##### Member
Sorry, I'm actually mistaking my terminology. It wasn't the partial fractions I needed assistance with, regardless I've worked out the problem I had before so this is redundant. Thanks for those that did post assistance, however. As an aside I'm self-taught so references to schooling timeframes are lost on me. I've learnt all I've learnt at my own pace and over a shorter time period than most.

So, are we arriving at a conclusion that going into hyperbolic tan inverses are an unneccesary step here? I wouldn't technically be wrong in doing it this way? (This is what I had attempted initially). Is it always better to use partial fractions when the denominator factorises thusly?

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MHB Math Helper

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#### chisigma

##### Well-known member
The partial fractions method is excellent in all cases... except when the denominator has complex roots. I wonder why nobody has examined the [much better...] procedure used by GreenGoblin...

$\displaystyle \frac{1}{x^{2}+x-6}= -\frac{1}{\frac{25}{4}-(x+\frac{1}{2})^{2}}$ (1)

... after that we have the definite integral 'reported in all manuals'...

$\displaystyle \int \frac{d x}{a^{2}-x^{2}}= \frac{1}{a}\ \tanh^{-1} \frac{x}{a} +c$ (2)

The advantage of the procedure is evident if we have to integrate $\displaystyle f(x)= \frac{1}{x^{2}+x+\frac{13}{2}}$. In this case the partial fraction method is difficult to apply but proceeding 'like GreenGoblin' we obtain...

$\displaystyle \frac{1}{x^{2}+x + \frac{13}{2}}= \frac{1}{\frac{25}{4}+(x+\frac{1}{2})^{2}}$ (3)

... and now the definite integral 'reported in all manuals' is...

$\displaystyle \int \frac{d x}{a^{2}+x^{2}}= \frac{1}{a}\ \tan^{-1} \frac{x}{a} + c$ (4)

Kind regards

$\chi$ $\sigma$

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#### GreenGoblin

##### Member
I appreciate the discussion, chisigma, I appreciate you take the time to consider the problem.

I do note that CaptainBlack is correct in that the partial fraction method is relatively straight forward but I also don't see that completing a square is that difficult. However, would it suffice to leave the solution in the form of an inverse (hyperbolic) tangent, or would it be a requirement to express this in the form of logarithms? (In which case the partial fraction method gets you there 'automatically').

Also one other pointer I noticed when looking at this problem is that, the derivative of inverse hyperbolic tangent and inverse hyperbolic cotangent are... identical... so, you could integrate this to inverse hyperbolic cotangent, correct or not? However, I note the identity for these in terms of logs are not identical. This I find confusing, can someone explain how this is possible please?

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#### CaptainBlack

##### Well-known member
I appreciate the discussion, chisigma, I appreciate you take the time to consider the problem.

I do note that CaptainBlack is correct in that the partial fraction method is relatively straight forward but I also don't see that completing a square is that difficult. However, would it suffice to leave the solution in the form of an inverse (hyperbolic) tangent, or would it be a requirement to express this in the form of logarithms? (In which case the partial fraction method gets you there 'automatically').
It was not completing the square in principle that I was complaining about but the was that Wolfram Alpha explained it.

CB

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