সৃজনশীলঃ ‘f(x)=x(x−1)(x2+1);g(x)=25−x2…’ রাজশাহী কলেজ ২০২৫ – যোগজীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

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

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

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

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

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

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

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

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

$=\frac{x^2}{2}{\cot }^{-1}(x-1)+\frac{1}{2}x+\frac{1}{2}\ln |x^2-2x+2|+c[\because \int \frac{f^{\prime }\left(x\right)}{f\left(x\right)}dx=\ln |(x)|+c]$

নির্ণেয় যোজিত ফল : $\frac{x^2}{2}{\cot }^{-1}(x-1)+\frac{1}{2}x+\frac{1}{2}\ln |x^2-2x+2|+c$.

${\int }_0^{5}g\left(x\right)dx={\int }_0^5\sqrt{25-x^2}dx={\int }_0^5\sqrt{5^2-x^2}dx$

$={[\frac{x\sqrt{5^2-x^2}}{2}+\frac{5^2}{2}{\sin }^{-1}\frac{x}{5}]}_0^5$

$[\because \int \sqrt{a^2-x^2}dx=\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}+c]$

$=(\frac{5\sqrt{25-25}}{2}+\frac{25}{2}{\sin }^{-1}\frac{5}{5})-(0+0)$

$=0+\frac{25}{2}\cdot \frac{\pi }{2}=\frac{25}{4}\pi$

$\therefore {\int }_0^{5}g\left(x\right)dx=\frac{25}{4}\pi .$