Question

If \[\int {\dfrac{{dx}}{{\left[ {{x^4} + {x^3}} \right]}}} = \left[ {\dfrac{A}{{{x^2}}}} \right] + \left( {\dfrac{B}{x}} \right) + \log \left| {\dfrac{x}{{(x + 1)}}} \right| + c\], then
A. \[A = 1/2,B = 1\]
B. \[A = 1,B = 1/2\]
C. \[A = - (1/2),B = 1\]
D. \[A = 1,B = 1\]

Answer 1

Hint: Integrate the value on the left-hand side and match it with the values on the right-hand side to find the values of constants A and B. Integrate the value on the left hand by dividing the denominator into separate parts using partial fractions.

Formula used:
 We have used the formula of partial fraction that is given below
 \[\dfrac{1}{{{x^3}(x + 1)}} = \dfrac{A}{{x + 1}} + \dfrac{{B{x^2} + cx + D}}{{{x^3}}}\]
Complete step by step answer: We are given an equation that is \[\int {\dfrac{{dx}}{{\left[ {{x^4} + {x^3}} \right]}}} = \left[ {\dfrac{A}{{{x^2}}}} \right] + \left( {\dfrac{B}{x}} \right) + \log \left| {\dfrac{x}{{(x + 1)}}} \right| + c\]
Now we apply the formula of partial fraction in the given equation, we get
\[
  \dfrac{1}{{{x^3}(x + 1)}} = \dfrac{A}{{x + 1}} + \dfrac{{B{x^2} + cx + D}}{{{x^3}}} \\
  1 = A{x^3} + B{x^3} + C{x^2} + Dx + B{x^2} + Cx + D \\
  1 = (A + B){x^3} + (C + B){x^2} + (D + C)x + D \\
\]
The required equations are
\[
  A + B = 0 \\
  C + B = 0 \\
  D + C = 0 \\
  D = 1 \\
\]
So, the required values are
 \[
  A = - 1 \\
  B = 1 \\
  C = - 1 \\
  D = 1 \\
\]
Now, we substitute the values in the given equation
\[
  \int {\dfrac{{dx}}{{{x^3}(x + 1)}}} = \int {\left( { - \dfrac{1}{{x + 1}} + \dfrac{{{x^3} - x + 1}}{{{x^3}}}} \right)} dx \\
   = \int {\left( { - \dfrac{1}{{x + 1}} + \dfrac{1}{x} - \dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^3}}}} \right)} dx \\
   - \ln |x + 1| + \ln |x| + \dfrac{1}{x} - \dfrac{1}{{2{x^2}}} + c \\
\]
Therefore, the value of \[A = - \dfrac{1}{2},B = 1\]

So, option C is correct

Hence, the value of \[A = - \dfrac{1}{2},B = 1\].


Note: Many students made miscalculations while writing the values of A, B, C and D so make sure about the formula and also solve the question with the help of the formula. The result on the right-hand side is given in separate terms even though we have to integrate only a single term on the left-hand side, so students should be able to identify that we have to use partial fractions.