Math, asked by inavu, 4 months ago

plz answer this question guys​

Attachments:

Answers

Answered by vinnu81
2

hope u understand this answer .

Attachments:
Answered by Seafairy
17

Given :

\displaystyle{\sf  \int \log(2+x^2)dx}

Solution :

\displaystyle{\Rightarrow \sf \int \log (2+x^2)dx}\\

\displaystyle{\Rightarrow \sf \log (2+x^2)\int 1\:dx-\int\Bigg(\dfrac{d\:\log(2+x^2)}{dx}.\int \:1\:dx\Bigg)dx}\\

\displaystyle{\Rightarrow \sf \log(2+x^2)-x-\int\dfrac{1-2x}{2+x^2}\:xdx}\\

\displaystyle{\Rightarrow \sf x-\log(2+x^2)-\int \dfrac{2x^2}{x+x^2}\:dx}\\

\displaystyle{\Rightarrow \sf x\:\log(2+x^2)-2\int\dfrac{x^2+2-2}{2+x^2}\:dx}\\

\displaystyle{\Rightarrow \sf x\log \:x (2+x^2)-2\Bigg[\int \bigg(\int 1\:dx\bigg)-\int \dfrac{2}{2+x^2}\:dx\Bigg]}\\

\displaystyle{\Rightarrow \sf x\log\:(2+x^2)-\Bigg[x - \bigg(2\int\dfrac{1}{2+x}\bigg)dx\Bigg]}\\

\displaystyle{\Rightarrow \sf x\:\log(2+x^2)-\Bigg[x-2\bigg(\dfrac{1}{\sqrt{2}}\:\tan^{-1}\dfrac{x}{\sqrt{2}}\bigg)\Bigg]+C}\\

\displaystyle{\Rightarrow \sf x\:\log(2+x^2)-2x+2\sqrt{2}tan^{-1}\dfrac{x}{\sqrt{2}}+C}

Required Answer :

\displaystyle{\boxed{\boxed{ \sf \int \log(2+x^2)dx=x\:\log(2+x^2)-2x+2\sqrt{2}tan^{-1}\dfrac{x}{\sqrt{2}}+C}}}

Similar questions