The given function is \( \displaystyle \frac{x}{\sqrt{1-2 x}} \) and we need to find the integration of this function with respect to \( x \).
Solution:
\( \displaystyle \int \frac{x}{\sqrt{1 – 2x}} \, dx \)Let \( u = 1-2x \), then \( \displaystyle x = \frac{1-u}{2} \) and
\( \displaystyle \frac{du}{dx} = -2 \implies dx = -\frac{1}{2} du \\ \)Rewrite the integral in terms of \( u \), substitute \( x \) and \( dx \):
\( \begin{aligned} &\int \frac{1-u}{2} \cdot \frac{1}{\sqrt{u}} \cdot \left(\frac{-1}{2} \right)du \\ &= -\frac{1}{4} \int \frac{1–u}{\sqrt{u}} du \\ &= -\frac{1}{4} \int \left( u^{-1/2} \,- u^{1/2} \right) du \\ &= -\frac{1}{4} \left( 2u^{1/2} – \frac{2}{3} u^{3/2} \right) + C, \quad \text{C is a constant} \\ &= -\frac{1}{2} u^{1/2} + \frac{1}{6} u^{3/2} + C \\ &= -\frac{1}{6}u^{1/2} \left(3-u \right) + C \\ \end{aligned} \)Substitute Back \( u = 1-2x \):
\( \begin{aligned} &-\frac{1}{6}(1-2x)^{1/2} \left(3-1+2x \right) + C \\ &= -\frac{1}{6}(1-2x)^{1/2} \left(2+2x \right) + C \\ &= -\frac{1}{3}(1-2x)^{1/2} \left(1+x \right) + C \\ &= -\frac{1}{3} \sqrt{1-2x} \left(1+x \right) + C \\ &= -\frac{1}{3} (1 + x) \sqrt{1 – 2x} + C \\ \end{aligned} \)Thus, the integration of the given function is
\( \displaystyle \boxed{ \int \frac{x}{\sqrt{1 – 2x}} \, dx = -\frac{1}{3} (1 + x) \sqrt{1 – 2x} + C } \)Please let me know in the comments if you find any errors in this solution.