Welcome to our community

Be a part of something great, join today!

Integral - substitution method problem

Yankel

Active member
Jan 27, 2012
398
Hello all

I am working on this integral

\[\int \frac{x^{2}+1}{x^{4}+1}dx\]


Now, I have tried this way:

\[u=x^{2}+1\]

after I did:

\[\int \frac{x^{2}+1}{\left ( x^{2}+1 \right )\left ( x^{2}-1 \right )}dx\]

But I got stuck, I got:

\[\frac{1}{2}\cdot \int \frac{1}{u\sqrt{u-1}}dx\]

I thought of making another substitution, but I tried and failed. Help required :confused:
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
Hello all

I am working on this integral

\[\int \frac{x^{2}+1}{x^{4}+1}dx\]


Now, I have tried this way:

\[u=x^{2}+1\]

after I did:

\[\int \frac{x^{2}+1}{\left ( x^{2}+1 \right )\left ( x^{2}-1 \right )}dx\]

But I got stuck, I got:

\[\frac{1}{2}\cdot \int \frac{1}{u\sqrt{u-1}}dx\]

I thought of making another substitution, but I tried and failed. Help required :confused:
Well first of all, \(\displaystyle \displaystyle \begin{align*} x^4 + 1 \end{align*}\) is NOT equal to \(\displaystyle \displaystyle \begin{align*} \left( x^2 + 1 \right) \left( x^2 - 1 \right) \end{align*}\).

I would try

\(\displaystyle \displaystyle \begin{align*} x^4 + 1 &= x^4 + 2x^2 + 1 - 2x^2 \\ &= \left( x^2 + 1 \right) ^2 - \left( \sqrt{2} \, x \right) ^2 \\ &= \left( x^2 - \sqrt{2}\, x + 1 \right) \left( x^2 + \sqrt{2}\,x + 1 \right) \end{align*}\)

and so

\(\displaystyle \displaystyle \begin{align*} \int{ \frac{x^2 + 1}{x^4 + 1}\,dx} &= \int{ \frac{x^2 - \sqrt{2}\,x + 1 + \sqrt{2}\,x}{ \left( x^2 - \sqrt{2}\,x + 1 \right) \left( x^2 + \sqrt{2}\,x + 1 \right) } \, dx} \\ &= \int{ \frac{1}{x^2 + \sqrt{2}\,x + 1} \,dx} + \sqrt{2} \int{ \frac{x}{ \left( x^2 - \sqrt{2}\,x + 1 \right) \left( x^2 + \sqrt{2}\,x + 1 \right) } \, dx} \\ &= \int{ \frac{1}{x^2 + \sqrt{2}\,x + \left( \frac{\sqrt{2}}{2} \right) ^2 - \left( \frac{\sqrt{2}}{2} \right) ^2 + 1 } \, dx} + \sqrt{2} \int{ \frac{x}{x^4 + 1} \, dx} \\ &= \int{ \frac{1}{ \left( x + \frac{\sqrt{2}}{2} \right) ^2 + \frac{1}{2} } \, dx} + \frac{\sqrt{2}}{2} \int{ \frac{2x}{ \left( x^2 \right) ^2 + 1 }\, dx} \end{align*}\)

Now in the first integral, let \(\displaystyle \displaystyle \begin{align*} x + \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2}\tan{(\theta)} \implies dx = \frac{\sqrt{2}}{2}\sec^2{(\theta)}\,d\theta \end{align*}\) and in the second integral let \(\displaystyle \displaystyle \begin{align*} x^2 = \tan{(\phi)} \implies 2x\,dx = \sec^2{(\phi)}\,d\phi \end{align*}\) and the integrals become

\(\displaystyle \displaystyle \begin{align*} \int{ \frac{1}{\left( x + \frac{\sqrt{2}}{2} \right) ^2 + \frac{1}{2} } \, dx} + \frac{\sqrt{2}}{2} \int{ \frac{2x}{ \left( x^2 \right) ^2 + 1 } \, dx} &= \int{ \frac{1}{ \left[ \frac{\sqrt{2}}{2} \tan{(\theta)} \right] ^2 + \frac{1}{2} } \cdot \frac{\sqrt{2}}{2}\sec^2{(\theta)}\,d\theta } + \frac{\sqrt{2}}{2} \int{ \frac{\sec^2{(\phi)}}{ \tan^2{(\phi)} + 1 } \, d\phi } \\ &= \frac{\sqrt{2}}{2} \int{ \frac{\sec^2{(\theta)}}{ \frac{1}{2}\tan^2{(\theta)} + \frac{1}{2} } \, d\theta} + \frac{\sqrt{2}}{2} \int{ \frac{\sec^2{(\phi)}}{\sec^2{(\phi)}}\,d\phi} \\ &= \sqrt{2} \int{ \frac{\sec^2{(\theta)}}{ \tan^2{(\theta)} + 1 } \, d\theta} + \frac{\sqrt{2}}{2} \int{ 1 \, d\phi } \\ &= \sqrt{2} \int{ \frac{\sec^2{(\theta)}}{\sec^2{(\theta)}}\,d\theta} + \frac{\sqrt{2}}{2} \phi + C_1 \\ &= \sqrt{2} \int{ 1\,d\theta } + \frac{\sqrt{2}}{2} \arctan{ \left( x^2 \right) } + C_1 \\ &= \sqrt{2} \, \theta + C_2 + \frac{\sqrt{2}}{2}\arctan{ \left( x^2 \right) } + C_1 \\ &= \sqrt{2} \arctan{ \left( \sqrt{2}\, x + 1 \right) } + \frac{\sqrt{2}}{2} \arctan{ \left( x^2 \right) } + C \textrm{ where } C = C_1 + C_2 \end{align*}\)
 

Yankel

Active member
Jan 27, 2012
398
sorry, I thought it was minus when I tried, of course x^4+1 can't go the way I wrote it.

I have a basic question, how did you got to trigonometry ? did you move to polar system or something ? I don't get it. I mean, how did you know you need trigonometry in the first place ?

Thanks !
 

paulmdrdo

Active member
May 13, 2013
386
I would try this method but this is open for any correction

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

by letting

$\displaystyle u\,=\,x^2$
$\displaystyle du\,=\,2xdx$
$\displaystyle dx\,=\,\frac{du}{2x}$
$\displaystyle x\,=\,u^{\frac{1}{2}}$

can you continue..?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
For practice try

\(\displaystyle \int \frac{1}{1+x^4}\,dx\)

\(\displaystyle \int \frac{1}{\sqrt[4]{1+x^4}}\, dx\)

\(\displaystyle \int \frac{x^2}{1+x^4}\, dx\)

They are a little bit tougher .

[*] By the way I have a way to solve the integral using complex numbers , the answer is nasty .
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Using partial fractions, we may write:

\(\displaystyle \int\frac{x^2+1}{x^4+1}\,dx=\frac{1}{2}\int\frac{1}{x^2+\sqrt{2}x+1}+\frac{1}{x^2-\sqrt{2}x+1}\,dx\)

Completing the squares...

\(\displaystyle \int\frac{x^2+1}{x^4+1}\,dx=\frac{1}{2}\int\frac{1}{\left(x+\frac{1}{\sqrt{2}} \right)^2+\frac{1}{2}}+\frac{1}{\left(x-\frac{1}{\sqrt{2}} \right)^2+\frac{1}{2}}\,dx\)

Now, we may simply apply the formula:

\(\displaystyle \int\frac{du}{u^2+a^2}=\frac{1}{a}\tan^{-1}\left(\frac{u}{a} \right)+C\)

to get (after simplification):

\(\displaystyle \int\frac{x^2+1}{x^4+1}\,dx=\frac{1}{\sqrt{2}} \left(\tan^{-1}\left(\sqrt{2}x+1 \right)+ \tan^{-1}\left(\sqrt{2}x-1 \right) \right)+C\)
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,404
sorry, I thought it was minus when I tried, of course x^4+1 can't go the way I wrote it.

I have a basic question, how did you got to trigonometry ? did you move to polar system or something ? I don't get it. I mean, how did you know you need trigonometry in the first place ?

Thanks !
It's a method known as trigonometric substitution. I suggest you read what Mark has written in post #8 of this thread.