int log (1+x)^{1+x} ,dx= | vedclass

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(C) Let $I = \int \log (1+x)^{1+x} \,dx$.Using the property $\log(a^b) = b \log a$,we get:$I = \int (1+x) \log (1+x) \,dx$.Substitute $t = 1+x$,then $

Vedclass · Accepted Answer

$\frac{(1+x)^2}{2}\left[\log (1+x)-\frac{1}{2}\right]+c$,where $c$ is a constant of integration.

Vedclass · Answer

(C) Let $I = \int \log (1+x)^{1+x} \,dx$.Using the property $\log(a^b) = b \log a$,we get:$I = \int (1+x) \log (1+x) \,dx$.Substitute $t = 1+x$,then $dt = dx$.$I = \int t \log t \,dt$.Using integration by parts $\int u v \,dt = u \int v \,dt - \int (u' \int v \,dt) \,dt$,where $u = \log t$ and $v = t$:$I = \log t \cdot \frac{t^2}{2} - \int (\frac{1}{t} \cdot \frac{t^2}{2}) \,dt$.$I = \frac{t^2}{2} \log t - \frac{1}{2} \int t \,dt$.$I = \frac{t^2}{2} \log t - \frac{1}{2} \cdot \frac{t^2}{2} + c$.$I = \frac{t^2}{2} [\log t - \frac{1}{2}] + c$.Substituting $t = 1+x$ back:$I = \frac{(1+x)^2}{2} [\log (1+x) - \frac{1}{2}] + c$.

$\int \log (1+x)^{1+x} \,dx=$

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