The integral intleft(frac{2 x^3-3 x+5}{2 x^2} | vedclass

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Question

(B) Given the integral $I = \int \left( \frac{2x^3 - 3x + 5}{2x^2} \right) dx$. Dividing each term in the numerator by $2x^2$,we get: $I = \int \left(

Vedclass · Accepted Answer

$x > 0$

Vedclass · Answer

(B) Given the integral $I = \int \left( \frac{2x^3 - 3x + 5}{2x^2} \right) dx$. Dividing each term in the numerator by $2x^2$,we get: $I = \int \left( \frac{2x^3}{2x^2} - \frac{3x}{2x^2} + \frac{5}{2x^2} \right) dx = \int \left( x - \frac{3}{2x} + \frac{5}{2x^2} \right) dx$. Integrating term by term: $I = \frac{x^2}{2} - \frac{3}{2} \ln|x| - \frac{5}{2} \left( -\frac{1}{x} \right) + C = \frac{x^2}{2} - \frac{3}{2} \ln|x| + \frac{5}{2x} + C$. However,the expression $\ln(x)$ is only defined for $x > 0$. Therefore,the integral is valid for $x > 0$. Hence,option $B$ is correct.

The integral $\int\left(\frac{2 x^3-3 x+5}{2 x^2}\right) d x$ is valid for:

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