Indefinite Integrals
13.0 Integral of type $\int {{x^m}{{\left( {a + b{x^n}} \right)}^p}dx} $
13.0 Integral of type $\int {{x^m}{{\left( {a + b{x^n}} \right)}^p}dx} $
To solve this type of integral, depending on the value of constants $m$, $n$ and $p$ different methods are applied which are as follows:
(i) If $p \in N$, we expand ${(a + b{x^n})^p}$ using binomial theorem then multiply it with ${x^m}$ and integrate it.
(ii) If $p \in {I^ - }$ i.e., negative integer, then we substitute $x = {t^k}$ where $k$ is the L.C.M. of $m$ and $n$.
(iii) If $\frac{{m + 1}}{n}$ is an integer and $p = \frac{j}{k}$ i.e., a fraction, then we substitute $a + b{x^n} = {t^k}$ where $k$ is the denominator of the fraction $p$.
(iv) If $\frac{{m + 1}}{n} + p$ is an integer and $p = \frac{j}{k}$ i.e., a fraction, then we substitute $a + b{x^n} = {t^k}{x^n}$ where $k$ is the denominator of the fraction $p$.
Question 12. Evaluate $$\int {{x^{ - \frac{2}{3}}}{{(1 + {x^{\frac{1}{3}}})}^{\frac{1}{2}}}} dx$$
Solution: On comparing it with $$\int {{x^m}{{\left( {a + b{x^n}} \right)}^p}dx} $$, we get
$m = - \frac{2}{3},n = \frac{1}{3},p = \frac{j}{k} = \frac{1}{2}$ and $\frac{{m + 1}}{n} = \frac{{ - \frac{2}{3} + 1}}{{\frac{1}{3}}} = 1$. Therefore, we put $$1 + {x^{\frac{1}{3}}} = {t^k} = {t^2} \Rightarrow \frac{1}{{3{x^{\frac{2}{3}}}}}dx = 2tdt$$$$\begin{equation} \begin{aligned} I = \int {\frac{{t.6t}}{1}dt} = 6\int {{t^2}dt} \\ I = 2{t^3} + C \\ I = 2{\left( {1 + {x^{\frac{1}{3}}}} \right)^{\frac{3}{2}}} + C \\\end{aligned} \end{equation} $$
