The primitive of f(x) = x cdot 2^{ln(x^2 + 1) | vedclass

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Question

(C) Let $I = \int x \cdot 2^{\ln(x^2 + 1)} dx$.Substitute $t = x^2 + 1$,then $dt = 2x dx$,which implies $x dx = \frac{dt}{2}$.The integral becomes $I

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$\frac{(x^2 + 1)^{\ln 2 + 1}}{2(\ln 2 + 1)} + C$

Vedclass · Answer

(C) Let $I = \int x \cdot 2^{\ln(x^2 + 1)} dx$.Substitute $t = x^2 + 1$,then $dt = 2x dx$,which implies $x dx = \frac{dt}{2}$.The integral becomes $I = \frac{1}{2} \int 2^{\ln t} dt$.Using the property $a^{\ln b} = b^{\ln a}$,we have $2^{\ln t} = t^{\ln 2}$.So,$I = \frac{1}{2} \int t^{\ln 2} dt$.Integrating with respect to $t$,we get $I = \frac{1}{2} \cdot \frac{t^{\ln 2 + 1}}{\ln 2 + 1} + C$.Substituting back $t = x^2 + 1$,we get $I = \frac{(x^2 + 1)^{\ln 2 + 1}}{2(\ln 2 + 1)} + C$.

The primitive of $f(x) = x \cdot 2^{\ln(x^2 + 1)}$ with respect to $x$ is:

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