-1-x2-x-x d x-A- log x-2 x-cB- log x2-x2-2-cC- log x2-x-x2-2-cD- log x2-x-x2-2-c

The correct option is C log x^2 x x^2/2+c∫(1 x)(2+x)/x dx=∫(2 2x+x x^2)/xdx =∫(2 x x^2)/xdx=∫2/xdx ∫ 1dx ∫ x dx =2log x x x^2/2+c =log x^2 x x^2/2+c ...

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Byju's Answer

Standard XII

Mathematics

First Fundamental Theorem of Calculus

∫1-x2+x/x d x...

Question

$\int \frac{(1 - x) (2 + x)}{x} d x =$

log x - 2 x + c

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log x^{2} - \frac{x^{2}}{2} + c

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log x^{2} - x - \frac{x^{2}}{2} + c

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log x^{2} + x + \frac{x^{2}}{2} + c

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Solution

The correct option is C $log x^{2} - x - \frac{x^{2}}{2} + c$
$\int \frac{(1 - x) (2 + x)}{x} d x = \int \frac{(2 - 2 x + x - x^{2})}{x} d x = \int \frac{(2 - x - x^{2})}{x} d x = \int \frac{2}{x} d x - \int 1 d x - \int x d x = 2 log x - x - \frac{x^{2}}{2} + c = log x^{2} - x - \frac{x^{2}}{2} + c$

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∫(1−x)(2+x)x dx=

The correct option is C logx2−x−x22+c∫(1−x)(2+x)x dx=∫(2−2x+x−x2)xdx=∫(2−x−x2)xdx=∫2xdx−∫1dx−∫xdx=2logx−x−x22+c=logx2−x−x22+c

$\int \frac{(1 - x) (2 + x)}{x} d x =$