Evaluate the following integrals :
(b) `int ( x-1)/( root(3) (x^2) ) dx,`

Answer

To evaluate the integral

\[ \int \frac{x - 1}{\sqrt[3]{x^2}} \, dx, \]

we can first simplify the integrand. We can rewrite the expression as:

\[ \frac{x - 1}{\sqrt[3]{x^2}} = \frac{x}{\sqrt[3]{x^2}} - \frac{1}{\sqrt[3]{x^2}}. \]

Now, we can express this in terms of powers of \(x\):

\[ \frac{x}{\sqrt[3]{x^2}} = x^{1 - \frac{2}{3}} = x^{\frac{1}{3}}, \]
and
\[ \frac{1}{\sqrt[3]{x^2}} = x^{-\frac{2}{3}}. \]

Thus, we can rewrite the integral as:

\[ \int \left( x^{\frac{1}{3}} - x^{-\frac{2}{3}} \right) \, dx. \]

Now we can integrate each term separately:

1. For the first term \(x^{\frac{1}{3}}\):

\[ \int x^{\frac{1}{3}} \, dx = \frac{x^{\frac{1}{3} + 1}}{\frac{1}{3} + 1} = \frac{x^{\frac{4}{3}}}{\frac{4}{3}} = \frac{3}{4} x^{\frac{4}{3}}. \]

2. For the second term \(-x^{-\frac{2}{3}}\):

\[ \int -x^{-\frac{2}{3}} \, dx = -\frac{x^{-\frac{2}{3} + 1}}{-\frac{2}{3} + 1} = -\frac{x^{\frac{1}{3}}}{\frac{1}{3}} = -3x^{\frac{1}{3}}. \]

Combining these results, we have:

\[ \int \frac{x - 1}{\sqrt[3]{x^2}} \, dx = \frac{3}{4} x^{\frac{4}{3}} - 3x^{\frac{1}{3}} + C, \]

where \(C\) is the constant of integration.

Thus, the final answer is:

\[ \int \frac{x - 1}{\sqrt[3]{x^2}} \, dx = \frac{3}{4} x^{\frac{4}{3}} - 3x^{\frac{1}{3}} + C. \]