Question

How do you integrate \[\dfrac{{{x^{\dfrac{1}{3}}}}}{{\left( {\left( {{x^{\dfrac{1}{3}}}} \right) - 1} \right)}}\]?

Answer 1

Hint: To solve this we will use a substitution method. We will first substitute the term in the denominator to be equal to some variable \[t\]. Then, we will try to write the numerator in terms of \[t\] by simplifying from the substitution. After that, we will find the value of \[dx\] by differentiating the expression we will obtain by substituting, in terms of \[t\] and \[dt\]. We will then have our expression in simplified form in terms of \[t\]. Integrating the whole expression with respect to \[t\] using the formulas:
\[\int {\left( {f\left( x \right) + g\left( x \right)} \right)} dx = \int {f\left( x \right)} dx + \int {g\left( x \right)} dx\]
\[\Rightarrow \int {{x^n}} dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c\], where \[c\] is a constant term.
\[\Rightarrow \int {\dfrac{1}{x}} dx = \ln |x| + c\], where \[c\] is a constant term.
\[\Rightarrow \int c dx = cx + k\], where \[k\] is a constant term.
We will then substitute the value of \[t\] back and obtain our answer in terms of \[x\].

Complete step by step answer:
We need to integrate \[\dfrac{{{x^{\dfrac{1}{3}}}}}{{\left( {\left( {{x^{\dfrac{1}{3}}}} \right) - 1} \right)}}\]. i.e. we need to find \[\int {\dfrac{{{x^{\dfrac{1}{3}}}}}{{\left( {\left( {{x^{\dfrac{1}{3}}}} \right) - 1} \right)}}} dx\].
Let us consider \[I = \int {\dfrac{{{x^{\dfrac{1}{3}}}}}{{\left( {\left( {{x^{\dfrac{1}{3}}}} \right) - 1} \right)}}} dx - - - - - - (1)\]
Let \[\left( {{x^{\dfrac{1}{3}}}} \right) - 1 = t - - - - - - (2)\]

Rearranging the terms in (2), we get
\[ \Rightarrow \left( {{x^{\dfrac{1}{3}}}} \right) = t + 1 - - - - - - (3)\]
Now, taking cube both the sides, we get
\[ \Rightarrow {\left( {{x^{\dfrac{1}{3}}}} \right)^3} = {\left( {t + 1} \right)^3}\]
Now, using the property \[{\left( {{a^m}} \right)^n} = {a^{mn}}\], we have
\[ \Rightarrow {\left( x \right)^{\dfrac{1}{3} \times 3}} = {\left( {t + 1} \right)^3}\]
\[ \Rightarrow {\left( x \right)^1} = {\left( {t + 1} \right)^3}\]
\[ \Rightarrow x = {\left( {t + 1} \right)^3}\]
Now, differentiating both sides with respect to \[t\], we get
\[ \Rightarrow \dfrac{{dx}}{{dt}} = \dfrac{d}{{dt}}\left( {{{\left( {t + 1} \right)}^3}} \right)\]

Now, using the property \[\dfrac{d}{{dx}}\left( {{{\left( {x + c} \right)}^n}} \right) = n{\left( {x + c} \right)^{n - 1}}\], where \[c\] is a constant term, we get
\[ \Rightarrow \dfrac{{dx}}{{dt}} = 3{\left( {t + 1} \right)^{3 - 1}}\]
\[ \Rightarrow \dfrac{{dx}}{{dt}} = 3{\left( {t + 1} \right)^2}\]
\[ \Rightarrow dx = 3{\left( {t + 1} \right)^2}dt - - - - - - (4)\]
Now, substituting (2), (3) and (4) in (1), we get
\[ \Rightarrow I = \int {\dfrac{{t + 1}}{t} \times 3{{\left( {t + 1} \right)}^2}} dt\]
Now, multiplying the terms in the numerator, we get
\[ \Rightarrow I = \int {\dfrac{{3{{\left( {t + 1} \right)}^3}}}{t}} dt\]
Now, using the Property \[\int {c\left( {f\left( x \right)} \right)} dx = c\int {\left( {f\left( x \right)} \right)} dx\], we have
\[ \Rightarrow I = 3\int {\dfrac{{{{\left( {t + 1} \right)}^3}}}{t}} dt\]

Simplifying the term in numerator using the formula \[{\left( {a + b} \right)^3} = {a^3} + {b^3} + 3{a^2}b + 3a{b^2}\], we get
\[ \Rightarrow I = 3\int {\dfrac{{{t^3} + {{\left( 1 \right)}^3} + 3{t^2}\left( 1 \right) + 3t{{\left( 1 \right)}^2}}}{t}} dt\]
As we know, \[{\left( 1 \right)^n} = 1\] for any \[n\]. So, our expression becomes
\[ \Rightarrow I = 3\int {\dfrac{{{t^3} + 1 + 3{t^2} + 3t}}{t}} dt\]
Now, separating the denominator with each term, we get
\[ \Rightarrow I = 3\int {\left( {\dfrac{{{t^3}}}{t} + \dfrac{1}{t} + \dfrac{{3{t^2}}}{t} + \dfrac{{3t}}{t}} \right)} dt\]
Now, cancelling out the terms, we get
\[ \Rightarrow I = 3\int {\left( {{t^2} + \dfrac{1}{t} + 3t + 3} \right)} dt\]

Now, using the property \[\int {\left( {f\left( x \right) + g\left( x \right)} \right)} dx = \int {f\left( x \right)} dx + \int {g\left( x \right)} dx\], we get
\[ \Rightarrow I = 3\left( {\int {{t^2}} dt + \int {\dfrac{1}{t}} dt + \int {3t} dt + \int 3 dt} \right)\]
Using \[\int {c\left( {f\left( x \right)} \right)} dx = c\int {\left( {f\left( x \right)} \right)} dx\] in the third term, we get
\[ \Rightarrow I = 3\left( {\int {{t^2}} dt + \int {\dfrac{1}{t}} dt + 3\int t dt + \int 3 dt} \right)\]
Now, using the formulas \[\int c dx = cx + k\], \[\int {{x^n}} dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c\] and \[\int {\dfrac{1}{x}} dx = \ln |x| + c\],where \[c\] is a constant, we get
\[ \Rightarrow I = 3\left( {\left( {\dfrac{{{t^{2 + 1}}}}{{2 + 1}} + {c_1}} \right) + \left( {\ln |t| + {c_2}} \right) + 3\left( {\dfrac{{{t^{1 + 1}}}}{{1 + 1}} + {c_3}} \right) + 3\left( {t + {c_4}} \right)} \right)\], where \[{c_1},{c_2},{c_3},{c_4}\] are constants

Solving the brackets, we have
\[ \Rightarrow I = 3\left( {\left( {\dfrac{{{t^3}}}{3} + {c_1}} \right) + \left( {\ln |t| + {c_2}} \right) + \left( {\dfrac{{3{t^2}}}{2} + 3{c_3}} \right) + \left( {3t + 3{c_4}} \right)} \right)\], where \[{c_1},{c_2},{c_3},{c_4}\] are constants
Now, combining all the constant terms, we get
\[ \Rightarrow I = 3\left( {\dfrac{{{t^3}}}{3} + \ln |t| + \dfrac{{3{t^2}}}{2} + 3t + \left( {{c_1} + {c_2} + 3{c_3} + 3{c_4}} \right)} \right)\], where \[{c_1},{c_2},{c_3},{c_4}\] are constants
Now, Opening the brackets, we get
\[ \Rightarrow I = \dfrac{{3{t^3}}}{3} + 3\ln |t| + \dfrac{{9{t^2}}}{2} + 9t + 3\left( {{c_1} + {c_2} + 3{c_3} + 3{c_4}} \right)\], where \[{c_1},{c_2},{c_3},{c_4}\] are constants
Letting \[3\left( {{c_1} + {c_2} + 3{c_3} + 3{c_4}} \right) = c\], we get
\[ \Rightarrow I = \dfrac{{3{t^3}}}{3} + 3\ln |t| + \dfrac{{9{t^2}}}{2} + 9t + c\], where \[c\] is a constant term.

Cancelling the term, we get
\[ \Rightarrow I = {t^3} + 3\ln |t| + \dfrac{{9{t^2}}}{2} + 9t + c\]
Now, substituting back the value of \[t\] from (2), we get
\[ \Rightarrow I = {\left( {{x^{\dfrac{1}{3}}} - 1} \right)^3} + 3\ln |{x^{\dfrac{1}{3}}} - 1| + \dfrac{{9{{\left( {{x^{\dfrac{1}{3}}} - 1} \right)}^2}}}{2} + 9\left( {{x^{\dfrac{1}{3}}} - 1} \right) + c\], where \[c\] is a constant term.
\[ \therefore I = {\left( {{x^{\dfrac{1}{3}}} - 1} \right)^3} + 3\ln |{x^{\dfrac{1}{3}}} - 1| + \dfrac{9}{2}{\left( {{x^{\dfrac{1}{3}}} - 1} \right)^2} + 9\left( {{x^{\dfrac{1}{3}}} - 1} \right) + c\], where \[c\] is a constant term.

Hence, \[\int {\dfrac{{{x^{\dfrac{1}{3}}}}}{{\left( {\left( {{x^{\dfrac{1}{3}}}} \right) - 1} \right)}}} dx = {\left( {{x^{\dfrac{1}{3}}} - 1} \right)^3} + 3\ln |{x^{\dfrac{1}{3}}} - 1| + \dfrac{9}{2}{\left( {{x^{\dfrac{1}{3}}} - 1} \right)^2} + 9\left( {{x^{\dfrac{1}{3}}} - 1} \right) + c\], where \[c\] is a constant term.

Note: We could have solved this problem by first adding and subtracting \[1\] in the numerator and then splitting the denominator to cancel out some terms and make the expression simple. After that, we will separately integrate both the terms and add them. To solve the integration, we take \[{x^{\dfrac{1}{3}}} = t\] and then make changes in the numerator and denominator and finding out \[dx\] in terms of \[t\] and then following the same procedure as we did in this question.