Question

The value of $\int \frac{x+5}{(x-2{)}^2}dx$ is :

• A. $\ln |(x-1)|-\frac{7}{(x-2)}+C$
  
• B. $\ln |(x-2)|-\frac{7}{(x-2)}+C$
• C. $\ln |(x-2)|-\frac{7}{(x-1)}+C$
• D. $\ln |(x-1)|-\frac{7}{(x-1)}+C$

Answer 1

The correct option is B

$\ln |(x-2)|-\frac{7}{(x-2)}+C$

We can see that the given expression is a proper fraction. Using partial fractions we'll split the given integrand as $\frac{A}{x-2}+\frac{B}{(x-2{)}^2}$

One thing we should notice here is that in the expression one of the terms has (x - 2) whereas another has $(x-2{)}^2$. Can you guess why? Well, it's because if we had taken (x - 2) in the denominator for both A and B ,on taking the LCM we would have never got ($x-2{)}^2$ in the denominator. Thus we'll not get the same expression. We know that we always have to split our expression as sum of terms such that the expression won't change.

Now let's take LCM and write the expression again-

$\frac{A}{x-2}+\frac{B}{(x-2{)}^2}=\frac{A(x-2)+B}{(x-2{)}^2}$

On comparing the coefficients of like powers of x, with the given expression, we can say-

A = 1

and B - 2A = 5

or B = 7

So, we'll have

$\int \frac{1}{(x-2)}+\frac{7}{(x-2{)}^2}dx$

$\ln |(x-2)|-\frac{7}{(x-2)}+C$