Question

Evaluate the following integral:
$\int {\dfrac{1}{{{x^{\dfrac{2}{3}}}\left( {{x^{\dfrac{1}{3}}} - 1} \right)}}} dx$.

Answer 1

Hint: Put ${x^{\dfrac{1}{3}}} = t$ in the integral to solve it and find the solution.

Complete step-by-step answer:

Given in the question the integral-
$\int {\dfrac{1}{{{x^{\dfrac{2}{3}}}\left( {{x^{\dfrac{1}{3}}} - 1} \right)}}} dx$
Let us put ${x^{\dfrac{1}{3}}} = t$.
Differentiate w.r.t x, we get-
${x^{\dfrac{1}{3}}} = t \\\ \Rightarrow \dfrac{1}{3}{x^{\dfrac{{ - 2}}{3}}}dx = dt - (1) \\\$
Put equation in the above integral, we get-
$\int {\dfrac{1}{{{x^{\dfrac{2}{3}}}\left( {t - 1} \right)}}} .\dfrac{{dt}}{{\dfrac{1}{3}{x^{\dfrac{{ - 2}}{3}}}}} \\\ \Rightarrow \int {\dfrac{1}{{{x^{\dfrac{2}{3}}}\left( {t - 1} \right)}}} .\dfrac{{3dt}}{{{x^{\dfrac{{ - 2}}{3}}}}} \\\ \Rightarrow \int {\dfrac{1}{{\left( {t - 1} \right)}}} .3dt \\\ \Rightarrow 3\log |t - 1| + C \\\$
Putting t = ${x^{\dfrac{1}{3}}}$, we get the value of the above integral as-
$3\log |{x^{\dfrac{1}{3}}} - 1| + C$ .
Hence, the answer is $3\log |{x^{\dfrac{1}{3}}} - 1| + C$.

Note: Whenever such types of questions appear then write the integral given in the question, substitute ${x^{\dfrac{1}{3}}} = t$ and differentiate it wrt to x. Then put ${x^{\dfrac{1}{3}}} = t$ in the given integral. The integral will transform into a simpler form as mentioned in the solution and then integrate it to find the value. Again, put t = ${x^{\dfrac{1}{3}}}$, in order to get the final answer.