Integration of $\int_{}^{}\frac{1}{x(1+x)^3}dx $

Question

we got this integration problem $$\int_{}^{}\frac{1}{x(1+x)^3}dx $$ it seems a fairly simple problem but what i am struggling with it is doing its partital fractions $\int_{}^{}\frac{1}{x(1+x)^3}dx $ =$\int_{}^{}\frac{a}{x}dx$ +$\int_{}^{}\frac{b}{(1+x)^3}dx$+ $\int_{}^{}\frac{c}{(1+x)^2}dx$ + $\int_{}^{}\frac{d}{1+x}dx$ now how to get values of $a,b,c,d$ ? it seems confusion by doing it by $a(1+x)^3+$ $bx+$ $cx(1+x)$ $+d(x(1+x)^2) =1$ is there any other way to break it into partial fractions and solve it ?

Answer 1

$$\frac1{x(x+1)^3}=\frac Ax+\frac B{x+1}+\frac C{(x+1)^2}+\frac D{(x+1)^3}\implies$$

$$1=A(x+1)^3+Bx(x+1)^2+Cx(x+1)+Dx$$

The above last expression is a polynomial identity and thus it is true for any value of $\;x\;$ , so substituting for example:

$$x=0\implies 1=A\cdot 1^3+0+0+0\implies A=1\\ x=-1\implies 1=0+0+0-D\implies D=-1$$

Now compare coefficients of corresponding powers of $\;x\;$ :

$$x^3\;\;\implies\;\;0=A+B\implies B=-A=-1\\x^2\;\;\implies\;\;0=3A+2B+C\implies C=-3+2=-1$$

Now check that what we got is correct...

Answer 2

You need to begin with the good partial fraction decomposition! That is, $$ \frac 1{x(1+x)^3} = \frac ax + \frac b{(1+x)} + \frac c{(1+x)^2} + \frac d{(1+x)^3}. $$ This is the correct way because the partial fraction decomposition works when you put irreducible factors individually in their separate fractions, together with $k$ terms if the irreducible factor appears to the $k^{\text{th}}$ power, each term being a different power of that irreducible factor. For instance, since $(1+x)^3$ is the irreducible factor $(1+x)$ (to the third power) of $x(1+x)^3$, it must appear three times, once for each power of $(1+x)$.

This gives you the equation $a(1+x)^3 + bx(1+x)^2 + cx(1+x) + dx = 1$. Expand the factors. You'll get $$ a + (3a + b + c + d)x + (3a + 2b + c)x^2 + (a + b) x^3 = 1. $$ By comparing the two polynomials' coefficients, it is straightforward that $a=1$ and $b = -a = -1$. Therefore $0 = 3a + 2b + c = 3 - 2 + c = 1+c$, hence $c=-1$ and $0 = 3a + b + c + d = 3 - 1 - 1 + d = 1+d$, hence $d = -1$. This gives you $$ \frac 1{x(1+x)^3} = \frac 1x + \frac {-1}{(1+x)} + \frac {-1}{(1+x)^2} + \frac {-1}{(1+x)^3}. $$ Hope that helps,

Answer 3

Ok, your equation is right: $$a(1 + x)^3 + bx + cx(1 + x) + d(1 + x)^2 = 1$$ Rewriting this: $$\begin{align} a(1 + 3x + 3x^2 + x^3) + bx + cx + cx^2 + dx(1 + 2x + x^2) & = 1 \\ a + 3ax + 3ax^2 + ax^3 + bx + cx+cx^2+dx+2dx^2+dx^3& = 1 \\ \end{align}$$ Grouping the terms by the powers of x: $$\begin{align} (a+d)x^3 + (3a+2d+c)x^2 + (3a+b+c+d)x + a = 1 \end{align}$$ Now as you can see, there are no $x^3, x^2, $ or $x$ terms on the right side of the equation. Therefore because the RHS (right hand side) has to equal the LHS (left hand side), we can make a system of equations for $a, b, c$ and $d$ $$\begin{align} x^3: a +d= 0 \\ x^2: 3a + 2d+c = 0 \\ x : 3a+b+c+2d = 0 \\ Constants: a = 1 \end{align}$$ Solving this we get $a = 1,$ $b = -1,$ $c = -1,$ $d = -1$ So the partial fraction decomposition is equal to: $$\frac{-1}{(1+x)^3} + \frac{-1}{(1+x)^2} + \frac{1}{x} + \frac{-1}{(1+x)}$$ The rest you can integrate.

Answer 4

Generalization:

$$I_n=\int\frac{dx}{x(1+x)^n}=\int\frac{(1+x-x)dx}{x(1+x)^n}=\int\frac{dx}{x(1+x)^{n-1}}-\int\frac{dx}{(1+x)^n}$$

Now using this, $$ \int\frac{dx}{(1+x)^n}=\int\frac{d(1+x)}{(1+x)^n}=\begin{cases} \ln|1+x| &\mbox{if } n=-1 \\ -\frac1{(n-1)(1+x)^{n-1}} & \text{ elsewhere } \end{cases} $$

$$\implies I_n=I_{n-1}-\begin{cases} \ln|1+x| &\mbox{if } n=-1 \\ -\frac1{(n-1)(1+x)^{n-1}} & \text{ elsewhere} \end{cases}$$

and $\displaystyle I_0=\int\frac{dx}x=\ln|x|+C$

Here we have $n=3$