Integration of $\int_{}^{}\frac{1}{x(1+x)^3}dx $ 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 ?
 A: $$\frac1{x(x+1)^3}=\frac Ax+\frac B{x+1}+\frac C{(x+1)^2}+\frac D{(x+1)^3}\implies$$
$$1=A(x+1)^3+Bx(x+1)^2+Cx(x+1)+Dx$$
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...
A: 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,
A: 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. 
A: 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 $n=3$
