Question

If m is a non-zero number and \[\int {\dfrac{{{x^{5m - 1}} + 2{x^{4m - 1}}}}{{{{\left( {{x^{2m}} + {x^m} + 1} \right)}^3}}}} dx = f\left( x \right) + c\], then \[f\left( x \right)\] is:
A) \[\dfrac{{{x^{5m}}}}{{2m{{\left( {{x^{2m}} + {x^m} + 1} \right)}^2}}}\]
B) \[\dfrac{{{x^{4m}}}}{{2m{{\left( {{x^{2m}} + {x^m} + 1} \right)}^2}}}\]
C) \[\dfrac{{2m\left( {{x^{5m}} + {x^{4m}}} \right)}}{{{{\left( {{x^{2m}} + {x^m} + 1} \right)}^2}}}\]
D) \[\dfrac{{\left( {{x^{5m}} - {x^{4m}}} \right)}}{{2m{{\left( {{x^{2m}} + {x^m} + 1} \right)}^2}}}\]

Answer 1

Hint: Here first we will simplify the given expression and then use the substitution and simple integration and then compare the function so obtained with the right hand side of the given equation to get the desired answer.
The general integral formula is:-
\[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C} \]

Complete step-by-step answer:
Let us consider the left hand side of the given equation.
Let,
\[I = \int {\dfrac{{{x^{5m - 1}} + 2{x^{4m - 1}}}}{{{{\left( {{x^{2m}} + {x^m} + 1} \right)}^3}}}} dx\]
Simplifying it further and taking \[{x^{2m}}\] common in denominator we get:-
\[I = \int {\dfrac{{{x^{5m - 1}} + 2{x^{4m - 1}}}}{{{{\left( {{x^{2m}}\left( {1 + {x^{ - m}} + {x^{ - 2m}}} \right)} \right)}^3}}}} dx\]
Simplifying it further we get:-
\[I = \int {\dfrac{{{x^{5m - 1}} + 2{x^{4m - 1}}}}{{{x^{6m}}{{\left( {1 + {x^{ - m}} + {x^{ - 2m}}} \right)}^3}}}} dx\]
Now dividing both the numerator and the denominator by \[{x^{6m}}\] we get:-
\[I = \int {\dfrac{{\dfrac{{{x^{5m - 1}} + 2{x^{4m - 1}}}}{{{x^{6m}}}}}}{{\dfrac{{{x^{6m}}{{\left( {1 + {x^{ - m}} + {x^{ - 2m}}} \right)}^3}}}{{{x^{6m}}}}}}} dx\]
Simplifying it further we get:-
\[
  I = \int {\dfrac{{\dfrac{{{x^{5m - 1}}}}{{{x^{6m}}}} + 2\dfrac{{{x^{4m - 1}}}}{{{x^{6m}}}}}}{{{{\left( {1 + {x^{ - m}} + {x^{ - 2m}}} \right)}^3}}}} dx \\
   \Rightarrow I = \int {\dfrac{{{x^{ - m - 1}} + 2{x^{ - 2m - 1}}}}{{{{\left( {1 + {x^{ - m}} + {x^{ - 2m}}} \right)}^3}}}} dx...................................\left( 1 \right) \\
 \]
Now let,
\[1 + {x^{ - m}} + {x^{ - 2m}} = t\]…………………………………..(2)
Differentiating both the sides we get:-
\[\dfrac{d}{{dx}}\left( {1 + {x^{ - m}} + {x^{ - 2m}}} \right) = \dfrac{d}{{dx}}\left( t \right)\]
Simplifying it further we get:-
\[\dfrac{d}{{dx}}\left( 1 \right) + \dfrac{d}{{dx}}\left( {{x^{ - m}}} \right) + \dfrac{d}{{dx}}\left( {{x^{ - 2m}}} \right) = \dfrac{{dt}}{{dx}}\]
Now we know that:-
\[\dfrac{d}{{dt}}\left( {{x^n}} \right) = n{x^{n - 1}}\]
Also, the differentiation of a constant term is zero.
Hence applying this formula we get:-
\[0 - m\left( {{x^{ - m - 1}}} \right) - 2m\left( {{x^{ - 2m - 1}}} \right) = \dfrac{{dt}}{{dx}}\]
Simplifying it further we get:-
\[
   - m\left[ {{x^{ - m - 1}} + 2{x^{ - 2m - 1}}} \right] = \dfrac{{dt}}{{dx}} \\
   \Rightarrow - m\left[ {{x^{ - m - 1}} + 2{x^{ - 2m - 1}}} \right]dx = dt \\
 \]
Solving it further we get:-
\[\left[ {{x^{ - m - 1}} + 2{x^{ - 2m - 1}}} \right]dx = \dfrac{{ - dt}}{m}\]…………………………..(3)
Now substituting the respective values of equation 2 and equation 3 in equation 1 we get:-
\[
  I = \int {\dfrac{{\dfrac{{ - dt}}{m}}}{{{{\left( t \right)}^3}}}} \\
   \Rightarrow I = \dfrac{{ - 1}}{m}\int {\dfrac{{dt}}{{{t^3}}}} \\
 \]
Now applying the general integral formula i.e.
\[\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C} \]
We get:-
\[I = \dfrac{{ - 1}}{m}\left[ {\dfrac{{ - 1}}{{2{t^2}}}} \right] + C\]
Simplifying it further we get:-
\[I = \dfrac{1}{{2m{t^2}}} + C\]
Putting back the value of t we get:-
\[I = \dfrac{1}{{2m{{\left( {1 + {x^{ - m}} + {x^{ - 2m}}} \right)}^2}}} + C\]
Simplifying it further we get:-
\[I = \dfrac{1}{{2m{{\left( {1 + \dfrac{1}{{{x^m}}} + \dfrac{1}{{{x^{2m}}}}} \right)}^2}}} + C\]
Taking the LCM we get:-
\[I = \dfrac{1}{{2m{{\left( {\dfrac{{{x^{2m}} + {x^m} + 1}}{{{x^{2m}}}}} \right)}^2}}} + C\]
On simplifying we get:-
\[
  I = \dfrac{{{{\left( {{x^{2m}}} \right)}^2}}}{{2m{{\left( {{x^{2m}} + {x^m} + 1} \right)}^2}}} + C \\
   \Rightarrow I = \dfrac{{{x^{4m}}}}{{2m{{\left( {{x^{2m}} + {x^m} + 1} \right)}^2}}} + C \\
 \]
Hence we got:
\[LHS = \dfrac{{{x^{4m}}}}{{2m{{\left( {{x^{2m}} + {x^m} + 1} \right)}^2}}} + C\]
Now equating it with right hand side we get:-
\[\dfrac{{{x^{4m}}}}{{2m{{\left( {{x^{2m}} + {x^m} + 1} \right)}^2}}} + C = f\left( x \right) + C\]
Hence on equating we get:-
\[f\left( x \right) = \dfrac{2}{{m{{\left( {1 + {x^{ - m}} + {x^{ - 2m}}} \right)}^2}}}\]
\[f\left( x \right) = \dfrac{{{x^{4m}}}}{{2m{{\left( {{x^{2m}} + {x^m} + 1} \right)}^2}}}\]

Hence option B is the correct option.

Note: In such questions we need to simplify the expression first and then make the required substitutions as we cannot directly proceed to find the integral of the function.
Substitution done should be such that all the terms in the integral are converted into new variables.
Students might forget to put the back value of t in the expression so they should take care of that.
Also, never forget to add the constant of integration after integrating the given expression.