Mathematics

The value of the integers $$\int\limits_0^2 {\dfrac{{\log ({x^2} + 2)}}{{{{\left( {x + 2} \right)}^2}}}dx\,is:} $$ 


ANSWER

$$\frac{{\sqrt 2 }}{3}\tan {\,^{ - 1}}\sqrt 2 + \frac{5}{{12}}\log \,2\, - \frac{1}{4}\log \,3$$


SOLUTION
$$I=\displaystyle\int^2_0\dfrac{log(x^2+2)}{(x+2)^2}$$
Integrating by parts by $$log(x^2+2)$$ as first function and $$\dfrac{1}{(x+2)^2}$$ as second function
$$\Rightarrow log(x^2+2)\displaystyle\int \dfrac{1}{(x+2)^2}dx+\displaystyle\int \dfrac{x}{x^2+2}\times \dfrac{1}{(x+2)}dx$$
Now, $$I_1=\displaystyle\int \dfrac{x}{(x^2+2)(x+2)}=A$$
Let $$\dfrac{x}{(x^2+2)(x+2)}=\dfrac{A}{x^2+2}+\dfrac{B}{x+2}$$
$$\dfrac{x}{(x^2+2)(x+2)}=\dfrac{Ax+2A+Bx^2+2B}{(x^2+2)(x+2)}$$
On comparing we get $$A=\dfrac{1}{3}, B=\dfrac{-1}{3}$$
$$\therefore I_1=\displaystyle\int \dfrac{x}{(x^2+2)(x+2)}=\displaystyle\int \left(\dfrac{1}{3(x^2+2)}-\dfrac{1}{3(x+2)}\right)dx$$
$$=\dfrac{2}{3}\tan^{-1}\dfrac{x}{\sqrt{2}}-\dfrac{1}{3}log|x+2|$$
Now, $$I=\left[\dfrac{-log (x^2+2)}{x+2}+\dfrac{2}{3}\tan^{-1}\dfrac{x}{\sqrt{2}}-log|x+2|\right]^2_0$$
$$=-\dfrac{log 6}{4}+\dfrac{2}{3}\tan^{-1}(\sqrt{2})-log 4+\dfrac{log 2}{2}-log 2$$
$$=\dfrac{z}{3}\tan^{-1}\sqrt{2}-\dfrac{5}{2}log^2-\dfrac{log 6}{4}$$
$$=\dfrac{2}{3}\tan^{-1}\sqrt{2}-\dfrac{5}{2}log^2-\dfrac{1}{4}[log^3+log^2]$$
$$=\dfrac{2}{3}\tan^{-1}\sqrt{2}-\dfrac{1}{4}log^3-\dfrac{11}{2}log^2$$.
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Single Correct Medium Published on 17th 09, 2020
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