Question

$\ln \left(1+x{e}^x\right)$

Short answer

$\ln \left(1+x{e}^x\right)=x+\frac{x^2}{2}-\frac{x^3}{6}-\frac{x^4}{12}+\dots$

Step-by-step solution

Given information

The Maclaurin expansion of the function$\ln \left(1+x{e}^x\right)$ is to be evaluated.

Definition of Maclaurin series

The definition of the Maclaurin series is defined.

$\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots$

$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots$

$x{e}^x=x+x^2+\frac{x^3}{2}+\frac{x^4}{6}+\dots$

Part-A Step 3: Begin by stating the Maclaurin series

State the Maclaurin series.

$\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots$

$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\dots$

$x{e}^x=x+x^2+\frac{x^3}{2}+\frac{x^4}{6}+\dots$

Part-B Step 3: Use the Maclaurin series to evaluate

Replace $x$with $\left[x+x^2+\frac{x^3}{2}+\frac{x^4}{6}+\dots \right]$in the first equation.

$\ln \left(1+x{e}^x\right)=\left[x+x^2+\frac{x^3}{2}+\frac{x^4}{6}+\dots \right]-\frac{{\left[x+x^2+\frac{x^3}{2}+\dots \right]}^2}{2}+\frac{{\left[x+x^2+\dots \right]}^3}{3}-\frac{[x+\dots {]}^4}{4}+\dots$

$\ln \left(1+x{e}^x\right)=\frac{24x+12x^2-4x^3-2x^4}{24}+\dots$

$\ln \left(1+x{e}^x\right)=x+\frac{x^2}{2}-\frac{x^3}{6}-\frac{x^4}{12}+\dots$

Thus, the final answer is$\ln \left(1+x{e}^x\right)=x+\frac{x^2}{2}-\frac{x^3}{6}-\frac{x^4}{12}+\dots$