Question

$\sqrt{1+\ln (1+x)}$

Short answer

$\sqrt{1+\ln (1+x)}=1+\frac{x}{2}-\frac{3x^2}{8}+\frac{17x^3}{48}+\dots$

Step-by-step solution

Given information

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

Definition of Maclaurin series

The definition of the Maclaurin series is defined.

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

$(1+x{)}^{1/2}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}+\dots$

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

State the Maclaurin series.

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

$(1+x{)}^{1/2}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}+\dots$

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

Substitute $x$with$\left[x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots \right]$in the second series and evaluate.

$\sqrt{1+\ln (1+x)}=1+\frac{\left[x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\dots \right]}{2}-\frac{{\left[x-\frac{x^2}{2}+\dots \right]}^2}{8}+\frac{[x+\dots {]}^3}{16}+\dots$

$\sqrt{1+\ln (1+x)}=1+\frac{\left(48x-36x^2+34x^3+\dots \right)}{96}$

role="math" localid="1657367767479" $\sqrt{1+\ln (1+x)}=1+\frac{x}{2}-\frac{3x^2}{8}+\frac{17x^3}{48}+\dots$

Thus, the final answer is$\sqrt{1+\ln (1+x)}=1+\frac{x}{2}-\frac{3x^2}{8}+\frac{17x^3}{48}+\dots$