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 ?

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,
