$I_1=\int f(x) d x$ and $I_2=\int_0^1 f(x) d x$ where $f(x)=x^2 \ln \left(1-x^2\right)$, then:

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Professor PC Thomas Classes · Answer

$ln(I-x)=-\{x+\frac{x^{2}}{2}+\frac{x^{3}}{3}+\frac{x^{4}}{4}+.......\}$ Replace x by $x^{2}$ $ln(1-x^{2})=-\{x^{2}+\frac{x^{4}}{2}+\frac{x^{6}}{3}+\frac{x^{8}}{4}+...\}$ $ln(1-x^{2})=-\{x^{4}+\frac{x^{6}}{2}+\frac{x^{8}}{3}+\frac{x^{8}}{4}+.......\}$ $x^{2}ln(1-x^{2})=-\{x^{4}+\frac{x^{6}}{2}+\frac{x^{8}}{3}+\frac{x^{10}}{4}+......\}$ $\int x^{2}ln(1-x^{2})dx=-\{\frac{x^{5}}{1.5}+\frac{x^{7}}{2.7}+\frac{x^{9}}{2.7}+\frac{x^{9}}{3.9}+\frac{x^{11}}{4.11}......\}$ Applying I.B.P with limits 0 to 1 for LHS $(\frac{x^{3}}{3}ln(1-x^{2}))_{0}^{1}-\int_{0}^{1}\frac{x^{3}}{3}\cdot\frac{(-2x)}{(1-x^{2})}dx$ $[\frac{x^{3}}{3}ln(1-x^{2})]_{0}^{1}+\frac{2}{3}(\frac{-x^{3}}{3}-x+\frac{1}{2}ln|\frac{1+x}{1-x}|)_{0}^{1}$ $\frac{1}{3}ln~2+\frac{1}{3}ln~2-\frac{2}{3}-\frac{2}{9}+lim_{xightarrow1}(\frac{x^{3}}{3}-\frac{1}{3})ln(1-x)$ $	herefore lim_{xightarrow1}(x^{3}-1)ln(1-x)=0$ $So\frac{2}{3}ln~2-\frac{8}{9}=RHS$ $:\frac{1}{1.5}+\frac{1}{2.7}+\frac{1}{3.9}+.......=\frac{2}{3}ln~2-\frac{8}{9}$