# Math Help - Integration using partial fractions.

1. ## Integration using partial fractions.

could someone help me

(x+5)dx/(x^2+x+1)

and this

(1/sqrt(2)*x+1)dx/(x^2-x*sqrt(2) +1)

2. Originally Posted by jo-ja
could someone help me

(x+5)dx/(x^2+x+1)
By completing the square: $x^2+x+1= x^2+x+ (1/4)- (1/4)+ 1= (x+1/2)^2+ 3/4$. That cannot be factored in terms of real numbers and so you cannot use "partial fractions". Make the substitution u= x+ 1/2 instead.

and this

(1/sqrt(2)*x+1)dx/(x^2-x*sqrt(2) +1)
Here, completing the square gives $x^2- x\sqrt{2}+ 1/2- 1/2+ 1= (x-\frac{\sqrt{2}}{2})^2+ 3/4$.

Again, you can't use partial fractions. Make the substitution $u= x-\frac{\sqrt{2}}{2}$ instead.

originaly was dx/(x^4+1) so then =>
dx/(x^2+x*sqrt(2)+1)*(x^2-x*sqrt(2)+1)
but then ....
Plz help

4. Consider

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

$= \int \frac{ 1 + 1/x^2 }{ x^2 + 1/x^2}~dx$

$= \int \frac{ 1+ 1/x^2}{ (x-1/x)^2 + 2} ~dx$

Sub $x - 1/x = t ~~\implies (1 + 1/x^2 )dx = dt$

the integral becomes

$\int \frac{dt}{ t^2 + 2}$

$= \frac{1}{\sqrt{2}} \tan^{-1}( \frac{x - 1/x}{\sqrt{2}}) + c$

similarly for $\int \frac{ x^2 - 1}{x^4 + 1}~dx$

$= - \frac{1}{\sqrt{2}} \tanh^{-1}( \frac{\sqrt{2}}{x + 1/x}) + c$

Finally , take sum of difference of them , what will you see then ?

5. I dont understand nothing coud someone plz solve mi this integral to the end

6. Originally Posted by jo-ja
I dont understand nothing coud someone plz solve mi this integral to the end

Here is the technique of resolving $\frac{1}{1 + x^n}$ into partial fractions .

$(1 + x^n ) = (1 +x ) ( 1 - ax + x^2 ) ( 1 - bx + x^2 ) ...$

( * the term " 1 + x " exists only for n is odd )
Then , " log " both sides

$\ln(1 + x^n) = \ln(x+1) + \ln(1 - ax + x^2) + \ln(1 - bx + x^2 ) + ...$

After that , differentiate both sides ( we call this " Logarithmic Differentiaion ! )

$\frac{ nx^{n-1}}{ 1 + x^n} = \frac{1}{1 + x} + \frac{ 2x - a}{1 - ax + x^2} + \frac{2x - b}{ 1 - bx + x^2} + ...$

Replace $x$ with $\frac{1}{x}$

$\frac{ n x }{ 1 + x^n } = \frac{x}{1 + x} + \frac{ 2x - ax^2 }{ 1 - ax + x^2 } + \frac{ 2x - bx^2 }{ 1 - bx + x^2 } + ...$

Divide both sides by $x$

$\frac{1}{ 1 + x^n } = \frac{1}{n} [ \frac{1}{ 1 + x } + \frac{ 2 - ax }{1 -ax + x^2} + \frac{ 2 -bx}{ 1 -bx + x^2} + ...$

ps : you should know what a ,b , c , d ... are . for this case ,
let $a = \sqrt{2} , b = -\sqrt{2}$

7. problem solved, tnx evryone, you are the best!!!!!