
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
163.5k+ views
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.
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.
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