int frac{x}{x + 2} dx is equal to | vedclass

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(D) Step $1$: Rewrite the numerator as $(x + 2) - 2$.Step $2$: Split the integral: $\int \frac{x + 2 - 2}{x + 2} dx = \int \left( \frac{x + 2}{x + 2}

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$x - 2 \log|x + 2| + c$

Vedclass · Answer

(D) Step $1$: Rewrite the numerator as $(x + 2) - 2$.Step $2$: Split the integral: $\int \frac{x + 2 - 2}{x + 2} dx = \int \left( \frac{x + 2}{x + 2} - \frac{2}{x + 2} \right) dx$.Step $3$: Simplify the expression: $\int (1 - \frac{2}{x + 2}) dx$.Step $4$: Integrate term by term: $\int 1 dx - 2 \int \frac{1}{x + 2} dx = x - 2 \log|x + 2| + c$.

$\int \frac{x}{x + 2} dx$ is equal to

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