Solution.

Here we need to make the substitution \(x = t^n\) where both \(\dfrac {n}{2}\) and \(\dfrac {n}{3}\) are integers.

Thus, the substitution is \(\hspace {0.2cm} x = t^6\hspace {0.2cm}\) so that \(\hspace {0.2cm} dx = 6t^5dt\)

\begin {align*} \implies \hspace {0.5cm} I & = \int \dfrac {6t^5}{\sqrt {t^6}- \sqrt [3]{t^6}}dt = 6\int \dfrac {t^5}{t^3 - t^2}dt\\\\ & = 6\int \dfrac {t^3}{t - 1}dt\\\\ & = 6\int \Big [t^2 + t + 1 + \dfrac {1}{t - 1}\Big ]dt\\\\\ & = 6\Bigg [\dfrac {1}{3}t^3 + \dfrac {1}{2}t^2 + t + \ln |t - 1|\Bigg ] + c\\\\ & = 6\Bigg [\frac {1}{3}x^{1/2} + \dfrac {1}{2}x^{1/3} + x^{1/6} + \ln \left |x^{1/6} - 1\right |\Bigg ] + c\\\\ \end {align*}

Solution.

Let \(\hspace {0.2cm} x + 1 = t^2\hspace {0.2cm}\), then \(\hspace {0.2cm} x = t^2 - 1\)

\(\therefore \hspace {0.4cm} x - 2 = t^2 - 3\hspace {0.5cm} \implies \hspace {0.5cm} dx = 2tdt\)

\begin {align*} I & = \int \dfrac {2t\hspace {0.1cm}dt}{\big (t^2 - 3\big ) t}\\\\ & = 2\int \dfrac {1}{\big (t - \sqrt {3}\big )\hspace {0.1cm} \big (t + \sqrt {3}\big )}\hspace {0.1cm}dt\\\\ & = \dfrac {2}{2\sqrt {3}}\int \Bigg (\dfrac {1}{t - \sqrt {3}} - \dfrac {1}{t + \sqrt {3}}\Bigg )\hspace {0.1cm}dt\\\\ & = \frac {1}{\sqrt {3}}\Big [\ln |t - \sqrt {3}| - \ln |t + \sqrt {3}|\Big ] + c\\\\ & = \dfrac {1}{\sqrt {3}}\ln \left |\dfrac {t - \sqrt {3}}{t + \sqrt {3}}\right | + c\\\\ & = \dfrac {1}{\sqrt {3}} \ln \left |\dfrac {\sqrt {x + 1} - \sqrt {3}}{\sqrt {x + 1} + \sqrt {3}}\right | + c\\\\ \end {align*}

Solution.

We complete the square of \(\hspace {0.3cm} x^2 - 6x + 13 = x^2 - 6x + 9 - 9 + 13 = \big (x - 3\big )^2 + 4\)

\begin {align*} \text {Thus}\hspace {0.5cm} I & = \int \dfrac {1}{\big (x - 3\big )^2 + 4}\hspace {0.1cm}dx\\\\ & = \dfrac {1}{2}\tan ^{-1}\dfrac {x - 3}{2} + c\\\\ \end {align*}