The integral int {frac{{xdx}}{{2 - {x^2} + sq | vedclass

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Question

(B) Let $I = \int \frac{x dx}{2 - x^2 + \sqrt{2 - x^2}}$.Substitute $t = \sqrt{2 - x^2}$. Then $t^2 = 2 - x^2$,which implies $2t dt = -2x dx$,or $x dx

Vedclass · Accepted Answer

$-\log \left| {1 + \sqrt {2 - {x^2}} } \right| + c$

Vedclass · Answer

(B) Let $I = \int \frac{x dx}{2 - x^2 + \sqrt{2 - x^2}}$.Substitute $t = \sqrt{2 - x^2}$. Then $t^2 = 2 - x^2$,which implies $2t dt = -2x dx$,or $x dx = -t dt$.Substituting these into the integral:$I = \int \frac{-t dt}{t^2 + t}$Simplify the integrand:$I = \int \frac{-t dt}{t(t + 1)} = -\int \frac{dt}{t + 1}$Integrating with respect to $t$:$I = -\log |t + 1| + c$Substituting back $t = \sqrt{2 - x^2}$:$I = -\log |\sqrt{2 - x^2} + 1| + c$.

The integral $\int {\frac{{xdx}}{{2 - {x^2} + \sqrt {2 - {x^2}} }}} $ equals

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