Question

The Maclaurin's expansion of In(1 + x2) is​

Answer 1

$ln(1+x^{2} )=$ $x^{2} -\frac{x^{4} }{2} +\frac{x^{6} }{3}........$ is the final Answer..

Step-1(Expansion of $ln(1+x)$ )

We know the expansion of  $ln(1+x)=x-\frac{x^{2} }{2}+\frac{x^{3} }{3} .........$ by Maclaurin Expansion...

Step-2( Calculation of Expansion of $ln(1+x^{2} )$)

Replacing $x^{2}$ in place of x in step 1 ,we will get the desire result..

Therefore the Expansion of $ln(1+x^{2} )=$ $x^{2} -\frac{x^{4} }{2}+\frac{x^{6} }{3} ....$