Question

Solution:

$$ \begin{aligned} \int \frac{(1-x)(2+x)}{x}dx &= \int \left(\frac{1 \cdot (2+x)}{x} - \frac{x \cdot (2+x)}{x}\right) dx \ &= \int \left(\frac{2}{x} + 1 - (2+x)\right) dx \ &= \int \left(\frac{2}{x} + 1 - 2 -x\right) dx \ &= \int \left(\frac{2}{x} - 1 -x\right) dx \ &= 2 \int \frac{1}{x} dx - \int 1 dx - \int x dx \ &= 2\ln|x| - x - \frac{x^2}{2} + c \ &= \ln x^2 + c - x - \frac{x^2}{2} + c \ &= \ln x^2 - x - \frac{x^2}{2} + c \end{aligned} $$

So, the correct answer is:

C) $\ln x^{2} - x - \frac{x^{2}}{2} + c$

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Answer

:

The correct option is C) $\log x^{2}-x-\frac{x^{2}}{2} + c$

Let's solve the integral: $$ \int \frac{(1-x)(2+x)}{x} , dx $$

First, expand the expression inside the integral: $$ \int \frac{(1-x)(2+x)}{x} , dx = \int \frac{2 + x - 2x - x^2}{x} , dx $$

Simplify the terms: $$ = \int \frac{2 - x - x^2}{x} , dx = \int \left( \frac{2}{x} - 1 - x \right) , dx $$

Now, integrate term by term: $$ = \int \frac{2}{x} , dx - \int 1 , dx - \int x , dx $$

Evaluate each integral separately: $$ = 2 \log x - x - \frac{x^2}{2} + c $$

Rewrite using logarithmic properties: $$ = \log x^2 - x - \frac{x^2}{2} + c $$

So, the solution is:

C) $\log x^{2} - x - \frac{x^{2}}{2} + c$


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