int log (2+x)^{2+x} , dx = | vedclass

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(A) Let $I = \int \log (2+x)^{2+x} \, dx$. Using the property $\log(a^b) = b \log a$,we get: $I = \int (2+x) \log(2+x) \, dx$. Let $u = 2+x$,then $du

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$\frac{(2+x)^2}{2} \log \left(\frac{2+x}{\sqrt{e}}\right)+c$,where $c$ is the constant of integration

Vedclass · Answer

(A) Let $I = \int \log (2+x)^{2+x} \, dx$. Using the property $\log(a^b) = b \log a$,we get: $I = \int (2+x) \log(2+x) \, dx$. Let $u = 2+x$,then $du = dx$. The integral becomes $I = \int u \log u \, du$. Using integration by parts $\int f(u)g'(u) \, du = f(u)g(u) - \int f'(u)g(u) \, du$,let $f(u) = \log u$ and $g'(u) = u$. Then $f'(u) = \frac{1}{u}$ and $g(u) = \frac{u^2}{2}$. $I = \frac{u^2}{2} \log u - \int \frac{1}{u} \cdot \frac{u^2}{2} \, du$ $I = \frac{u^2}{2} \log u - \frac{1}{2} \int u \, du$ $I = \frac{u^2}{2} \log u - \frac{1}{2} \cdot \frac{u^2}{2} + c$ $I = \frac{u^2}{2} \left( \log u - \frac{1}{2} \right) + c$ Since $\frac{1}{2} = \log \sqrt{e}$,we have: $I = \frac{u^2}{2} (\log u - \log \sqrt{e}) + c = \frac{u^2}{2} \log \left( \frac{u}{\sqrt{e}} \right) + c$. Substituting $u = 2+x$ back: $I = \frac{(2+x)^2}{2} \log \left( \frac{2+x}{\sqrt{e}} \right) + c$.

$\int \log (2+x)^{2+x} \, dx =$

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