Question

Resolve into a partial fraction $\dfrac{(2x-5)}{(x+3){{(x+1)}^{2}}}$.
(A) $\dfrac{11}{4(x+3)}-\dfrac{7}{2{{(x+1)}^{2}}}+\dfrac{11}{4(x+1)}$
(B) $-\dfrac{11}{4(x+3)}-\dfrac{7}{2{{(x+1)}^{2}}}+\dfrac{11}{4(x+1)}$
(C) $-\dfrac{11}{4(x+3)}+\dfrac{7}{2{{(x+1)}^{2}}}+\dfrac{11}{4(x+1)}$
(D) $\dfrac{11}{4(x+3)}+\dfrac{7}{2{{(x+1)}^{2}}}+\dfrac{11}{4(x+1)}$

Answer 1

Partial fraction is used to make the calculations of integration easier for us. Partial fractions can be of many types. We can have linear equations in the denominator, we can also have the quadratic equation in the denominator or we can even have the square root in the denominator and there are different methods to solve them.

Complete step by step answer:
In the above question, we have to resolve $\dfrac{(2x-5)}{(x+3){{(x+1)}^{2}}}$ into a partial fraction. So this fraction can be written as shown below.
$\dfrac{2x-5}{(x+3){{(x+1)}^{2}}}=\dfrac{A}{(x+3)}+\dfrac{B}{(x+1)}+\dfrac{C}{{{(x+1)}^{2}}}$……..eq(1)
Now we will take the LCM of the denominator which is as follows.
$\dfrac{2x-5}{(x+3){{(x+1)}^{2}}}=\dfrac{A{{(x+1)}^{2}}+B(x+1)(x+3)+C(x+3)}{(x+3){{(x+1)}^{2}}}$
Here we have to find the values of A, B, and C. So first to find out the value of A, we have to make the values of B and C is zero.
As the denominators of the above equations are equal, therefore they will cancel out each other and the below-written equation will be obtained.
$(2x-5)=A{{(x+1)}^{2}}+B(x+1)(x+3)+C(x+3)$………eq(2)
To find the value of A, B and C must be zero and to make them zero. We have to put the value of $x=-3$ in eq(2) and after putting the value, the following results will be obtained.

& (2x-5)=A{{(x+1)}^{2}}+B(x+1)(x+3)+C(x+3) \\\ & \Rightarrow (2(-3)-5)=A{{(-3+1)}^{2}}+B(-3+1)(-3+3)+C(-3+3) \\\ \end{aligned}$$ $$\Rightarrow A=\dfrac{-11}{4}$$ So the value of A is $$\dfrac{-11}{4}$$. We have to find the value of C. So for that, we have to make the value of A and B zero. So to make them zero we have to keep the value of $$x=-1$$ in eq(2) which is as follows. $$\begin{aligned} & (2x-5)=A{{(x+1)}^{2}}+B(x+1)(x+3)+C(x+3) \\\ & \Rightarrow (2(-1)-5)=A{{(-1+1)}^{2}}+B(-1+1)(-1+3)+C(-1+3) \\\ & \Rightarrow (-2-5)=2C \\\ & \Rightarrow C=\dfrac{-7}{2} \\\ \end{aligned}$$ So the value of C is $$\dfrac{-7}{2}$$. So to find the value to B, we have to use the value of A and C which is as follows. $$(2x-5)=A{{(x+1)}^{2}}+B(x+1)(x+3)+C(x+3)$$ On putting the value of $$x=0$$, we get the following results. $$\begin{aligned} & (2(0)-5)=A{{(0+1)}^{2}}+B(0+1)(0+3)+C(0+3) \\\ & \Rightarrow -5=A+3B+3C \\\ \end{aligned}$$ We know the values of A and C, So we will put them in the above equation. $$\begin{aligned} & -5=\dfrac{-11}{4}+3B+3\left( \dfrac{-7}{2} \right) \\\ & \Rightarrow -3B=\dfrac{-11}{4}-\dfrac{21}{2}+5 \\\ \end{aligned}$$ On taking the LCM of the denominator. $$\begin{aligned} & -3B=\dfrac{-11-42+20}{4} \\\ & \Rightarrow -3B=\dfrac{-33}{4} \\\ & \Rightarrow B=\dfrac{11}{4} \\\ \end{aligned}$$ So the value of B comes out to be $$\dfrac{11}{4}$$. On putting the value of A, B, and C in eq(1), we get the following result. $$\begin{aligned} & \dfrac{(2x-5)}{(x+3){{(x+1)}^{2}}}=\dfrac{A}{(x+3)}+\dfrac{B}{(x+1)}+\dfrac{C}{{{(x+1)}^{2}}} \\\ & \Rightarrow \dfrac{(2x-5)}{(x+3){{(x+1)}^{2}}}=-\dfrac{11}{4(x+3)}+\dfrac{11}{4(x+1)}-\dfrac{7}{2{{(x+1)}^{2}}} \\\ \end{aligned}$$ **So, the correct answer is “Option B”.** **Note:** Fraction can be of two types, proper fraction and improper fraction. If the degree of the numerator is less than the degree of the denominator in the fraction then it is a proper fraction but if the degree of the numerator is greater than the degree of the denominator then it is an improper fraction.