91Ó°ÊÓ

Let us consider the given function$f\left(x\right)=\ln (4+x^2)$

The maclaurin series for $g\left(x\right)=\ln (1+x)$is :

$\ln (1+x^2)=\underset{k=0}{\overset{∞}{∑}}\frac{{\left(-1\right)}^k}{k}{x}^k$

So,the maclaurin series for $f\left(x\right)=\ln (4+x^2)$Let us write the function as

$\ln (4+x^2)=\ln \left\{4\left[1+{\left(\frac{x}{2}\right)}^2\right]\right\}$

since, $\ln (a.b)=\ln a+\ln b$

Therefore,,

$\ln (4+x^2)=\ln 4+\ln \left[1+{\left(\frac{x}{2}\right)}^2\right]$

Now substitutexby${\left(\frac{x}{2}\right)}^2$in the maclaurin series of$g\left(x\right)=\ln (1+x)$to find the maclaurin series of$\ln \left[1+{\left(\frac{x}{2}\right)}^2\right]$

Thus,

$\ln (4+x^2)=\ln 4+\underset{k=1}{\overset{∞}{∑}}\frac{{\left(-1\right)}^k}{k}{\left(\frac{x}{2}\right)}^{2k}$

Inmplies that

$\ln (4+x^2)=\ln 4+\underset{k=1}{\overset{∞}{∑}}\frac{{\left(-1\right)}^k}{k.2^{2k}}.x^{2k}$

Let us consider the given function$F=∫f\left(x\right)$

Put the value of function$f\left(x\right)$

$$\begin{gathered} F\left(x\right)=∫\left[\ln 4+\underset{k=1}{\overset{∞}{∑}}\frac{{(-1)}^{k+1}}{k.2^{2k}}.x^{2k}\right]dx \\ =\left(\ln 4\right)∫dx+∫\left[\underset{k=1}{\overset{∞}{∑}}\frac{{(-1)}^{k+1}}{k.2^{2k}}.x^{2k}\right]dx \\ =(\ln 4)∫dx+\underset{k=1}{\overset{∞}{∑}}\frac{{(-1)}^{k+1}}{k.2^{2k}}∫x^{2k}dx \\ =(\ln 4)x+\underset{k=1}{\overset{∞}{∑}}\frac{{(-1)}^{k+1}}{k.2^{2k}}\frac{x^{2k+1}}{2k+1}+C \end{gathered}$$

Since,the initial condition is$F\left(0\right)=-2$

This implies that:

$C=-2$

Therefore,

$F\left(x\right)=(\ln 4)x+\underset{k=1}{\overset{∞}{∑}}\frac{{(-1)}^{k+1}}{k.2^{2k}}\frac{x^{2k+1}}{2k+1}-2$