$\int \log (1+x)^{1+x} \mathrm{~d} x=$

$\int \log (1+x)^{1+x} \mathrm{~d} x=$
  1. $(1+x)^2 \log (1+x)-\frac{1}{2}+\mathrm{c}$; where c is a constant of integration.
  2. $\frac{(1+x)^2}{2} \cdot \log (1+x)+\mathrm{c}$, where c is a constant of integration.
  3. $\frac{(1+x)^2}{2}\left[\log (1+x)-\frac{1}{2}\right]+\mathrm{c}$, where c is a constant of integration.
  4. $\frac{1+x}{2} \log (1+x)+\mathrm{c}$, where c is a constant of integration.

Solution

$\begin{aligned} & \text {Let } \mathrm{I}=\int \log (1+x)^{1+x} \mathrm{~d} x \\ & \mathrm{I}=\int(1+x) \cdot \log (1+x) \mathrm{d} x \\ & \text {Put } 1+x=\mathrm{t} \\ \therefore \quad & \mathrm{d} x=\mathrm{dt} \\ \therefore \quad & \mathrm{I}=\int \mathrm{t} \cdot \log \mathrm{t} \cdot \mathrm{dt}\end{aligned}$ $\begin{aligned} & =\log \mathrm{t} \int \mathrm{t} \mathrm{~dt}-\int\left(\frac{\mathrm{d}}{\mathrm{dt}} \log \mathrm{t} \cdot \int \mathrm{tdt}\right) \mathrm{dt} \\ & =\log \mathrm{t} \cdot \frac{\mathrm{t}^2}{2}-\int\left(\frac{1}{\mathrm{t}} \times \frac{\mathrm{t}^2}{2}\right) \mathrm{dt} \\ & =\log \mathrm{t} \cdot \frac{\mathrm{t}^2}{2}-\frac{1}{2} \int \mathrm{t} d \mathrm{dt} \\ & =\log \mathrm{t} \frac{\mathrm{t}^2}{2}-\frac{\mathrm{t}^2}{4}+\mathrm{c} \\ & =\mathrm{t}^2 \frac{\log \mathrm{t}}{2}-\frac{\mathrm{t}^2}{4}+\mathrm{c} \\ & =\frac{\mathrm{t}^2}{2}\left[\log \mathrm{t}-\frac{1}{2}\right]+\mathrm{c} \\ \mathrm{I} & =\frac{(1+x)^2}{2}\left[\log (1+x)-\frac{1}{2}\right]+\mathrm{c}\end{aligned}$

Asked in: MHT CET 2024 (02 May Shift 1)

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