91Ó°ÊÓ

Let $ u = \sqrt[3]{2x + 1} $. We will substitute this into the integral, but first, we need to find how dx is related to du. Differentiate both sides with respect to x: $ \frac{d u}{d x} = \frac{1}{3}(2x+1)^{-\frac{2}{3}} \cdot 2 $ Now we can express dx in terms of du: $ d x = \frac{1}{2}(2x+1)^{\frac{2}{3}} \cdot du $ Substitute $ u\( and \) d x$ into the integral: $ \int \frac{x^{2}-7 x+1}{u} \cdot \frac{1}{2}(2x+1)^{\frac{2}{3}} \cdot du $

Simplify the integral by cancelling terms: $ \int \frac{(x^{2}-7x+1)(2x+1)^{\frac{2}{3}}}{2u} \cdot du $ Now, we need to express the integral in terms of u. To do this, we can use the substitution $ u=\sqrt[3]{2x+1}$ to write x in terms of u: $ x = \frac{u^{3}-1}{2} $ Substitute x into the integral: $ \int \frac{(\frac{u^3 - 1}{2}^2 - 7\frac{u^3 - 1}{2} + 1)(2u^3-1)^{\frac{2}{3}}}{2u} \cdot du $

Now, we need to simplify the expression inside the integral before we can perform the integration: $ \int \frac{(u^6 -2u^3 + 1 -7u^3 +7 + 2u^2 - 2u^5)(2u^3-1)^{\frac{2}{3}}}{4u} \cdot du $ Combine like terms: $ \int \frac{(u^6-2u^5 -5u^3 + 2u^2 + 6)(2u^3-1)^{\frac{2}{3}}}{4u} \cdot du $

Unfortunately, integrating this expression directly is very difficult. However, we can notice that the expression inside the integral is a polynomial multiplied by \((2u^3 -1)^{2/3}\). We can look for patterns in the integral or apply a specialized integration technique like integration by parts, but none will provide a straightforward solution in this case. The given integral is quite challenging and does not have an elementary solution (i.e., a solution expressible in terms of elementary functions). However, if we still need to find the integral, we can resort to using numerical integration techniques or special functions. So, while an elementary solution could not be found, we made every effort to simplify the expression and understand the properties of the integral.