The integral of frac{x^2 - x}{x^3 - x^2 + x - | vedclass

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Question

(A) Let $I = \int \frac{x^2 - x}{x^3 - x^2 + x - 1} dx$. Factor the denominator: $x^3 - x^2 + x - 1 = x^2(x - 1) + 1(x - 1) = (x^2 + 1)(x - 1)$. Subst

Vedclass · Accepted Answer

$\frac{1}{2} \log (x^2 + 1) + c$

Vedclass · Answer

(A) Let $I = \int \frac{x^2 - x}{x^3 - x^2 + x - 1} dx$. Factor the denominator: $x^3 - x^2 + x - 1 = x^2(x - 1) + 1(x - 1) = (x^2 + 1)(x - 1)$. Substitute this into the integral: $I = \int \frac{x(x - 1)}{(x^2 + 1)(x - 1)} dx$. Cancel the common term $(x - 1)$: $I = \int \frac{x}{x^2 + 1} dx$. Multiply and divide by $2$: $I = \frac{1}{2} \int \frac{2x}{x^2 + 1} dx$. Let $u = x^2 + 1$,then $du = 2x dx$. Substituting these into the integral: $I = \frac{1}{2} \int \frac{1}{u} du = \frac{1}{2} \log |u| + c$. Since $x^2 + 1 > 0$ for all real $x$,we can write: $I = \frac{1}{2} \log (x^2 + 1) + c$.

The integral of $\frac{x^2 - x}{x^3 - x^2 + x - 1}$ with respect to $x$ is:

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