Question

$\int \frac{5xdx}{(1-x{)}^3}$ is equal to :

• A. $\frac{5}{2(x-1{)}^2}-\frac{5}{(x-1)}+C$
• B. $\frac{5}{2(x-1{)}^2}+\frac{5}{(x-1)}+C$
• C. $\frac{5}{3(x-1{)}^2}+\frac{5}{2(x-1)}+C$
• D. $\frac{5}{3(x-1{)}^2}-\frac{5}{2(x-1)}+C$
• E. $\frac{-5}{2(x-1{)}^2}+\frac{5}{(x-1)}+C$

Answer 1

Answer: (B)

$\frac{5}{2(x-1{)}^2}+\frac{5}{(x-1)}+C$

Let $I=\int \frac{5x}{(1-x{)}^3}dx$
Let $\frac{5x}{(1-x{)}^3}=\frac{A}{(1-x)}+\frac{B}{(1-x{)}^2}+\frac{C}{(1-x{)}^3}$
$\Rightarrow 5x=A(1-x{)}^2+B(1-x)+C$
$\Rightarrow 5x=A(1-2x+x^2)+B(1-x)+C$
On equating the coefficients of $x^2,x$ and constant terms, we get
$0=A,5=-2A-B,0=A+B+C$
Therefore, $5=-2\left(0\right)-B\Rightarrow B=-5$
and $0=0-5+C$
$\Rightarrow C=5$
Thus $\frac{5x}{(1-x{)}^3}=0+\frac{-5}{(1-x{)}^2}+\frac{5}{(1-x{)}^3}$
On integrating both sides, we get
$\int \frac{5x}{(1-x{)}^3}dx=\int -\frac{5}{(1-x{)}^2}dx+\int \frac{5}{(1-x{)}^3}dx$
$=\frac{-5(-1)}{-1(1-x{)}^1}+\frac{5(-1)}{-2(1-x{)}^2}$
$=-\frac{5}{1-x}+\frac{5}{2(1-x)}+C$
$=\frac{5}{2(x-1{)}^2}+\frac{5}{x-1}+C$