সৃজনশীলঃ ‘দৃশ্যকল্প : g(x)=cot−1⁡(x−1),f(x)=x…’ ময়মনসিংহ বোর্ড ২০২১ – যোগজীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

$f\left(x\right)=x$ এবং $g\left(x\right)={\cot }^{-1}(x-1)$

$\therefore \int f\left(x\right)g\left(x\right)\mathrm{dx}=\int x{\cot }^{-1}(x-1)$

$={\cot }^{-1}(x-1)\int xdx-\int \{\frac{d}{\mathrm{dx}}{\cot }^{-1}(x-1)\int \mathrm{xdx}\}\mathrm{dx}$

$={\cot }^{-1}(x-1)\frac{x^2}{2}-$

$\int \frac{-1}{1+(x-1{)}^2}\cdot \frac{d}{\mathrm{dx}}(x-1)\cdot \frac{x^2}{2}\mathrm{dx}$

$=\frac{x^2}{2}{\cot }^{-1}(x-1)+\int \frac{-1}{1+x^2-2x+1}\cdot \frac{x^2}{2}\mathrm{dx}$

$=\frac{x^2}{2}{\cot }^{-1}(x-1)+\frac{1}{2}\int \frac{x^2}{x^2-2x+2}\mathrm{dx}$

$=\frac{x^2}{2}{\cot }^{-1}(x-1)+\frac{1}{2}\int \frac{(x^2-2x+x)2x-2}{x^2-2x+2}\mathrm{dx}$

$=\frac{x^2}{2}{\cot }^{-1}(x-1)+\frac{1}{2}\int (1+\frac{2x-2}{x^2-2x+2})\mathrm{dx}$

$=\frac{x^2}{2}{\cot }^{-1}(x-1)+\frac{1}{2^x}+\frac{1}{2}\ln |x^2-2x+2|+c$

$[\because \int \frac{f\left(x\right)}{f\left(x\right)}\mathrm{dx}=\ln |f(x\left)|\right]$

নির্ণেয় যোজিত ফল :

$=\frac{x^2}{2}{\cot }^{-1}(x-1)+\frac{1}{2^x}+\frac{1}{2}\ln |x^2-2x+2|+c$

দেওয়া আছে,

$f\left(x\right)=x$

$\therefore \int \frac{f\left(x\right)}{\left\{f\right(x)-1\}\{f\left(x^2\right)+1\}}dx$

$=\int \frac{x}{(x-1)(x^2+1)}dx$

$=\int \frac{(x-1)+1}{(x-1)(x^2+1)}dx$

$=\int (\frac{1}{x^2+1}+\frac{1}{(x-1)(x^2+1)})dx$

$={\tan }^{-1}x+\int \frac{1}{(x-1)(x^2+1)}dx$

$={\tan }^{-1}x+\frac{1}{2}\int \{\frac{1}{(x-1)}-\frac{x+1}{x^2+1}\}dx$

$={\tan }^{-1}x+\frac{1}{2}\int \frac{dx}{x-1}-\frac{1}{2}\int \frac{x+1}{x^2+1}dx$

$={\tan }^{-1}x+\frac{1}{2}\ln |x-1|-\frac{1}{2}\int \frac{x+1}{x^2+1}dx$

$={\tan }^{-1}x+\frac{1}{2}\ln |x-1|-\frac{1}{2}\int \frac{x}{x^2+1}dx-\frac{1}{2}\int \frac{1}{x^2+1}dx$

$={\tan }^{-1}x+\frac{1}{2}\ln |x-1|-\frac{1}{4}\int \frac{2x}{x^2+1}dx-\frac{1}{2}{\tan }^{-1}x$

$=\frac{1}{2}{\tan }^{-1}x+\frac{1}{2}\ln |x-1|-\frac{1}{4}\ln |x^2+1|+c$

নির্ণেয় যোজিত ফল :

$\frac{1}{2}{\tan }^{-1}x+\frac{1}{2}\ln |x-1|-\frac{1}{4}\ln |x^2+1|+c$