সৃজনশীলঃ ‘f(x)=x^{2}$…’ রাজশাহী বোর্ড ২০২১ – যোগজীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

দেওয়া আছে, $f\left(x\right)=x^2$

$\therefore \int [\frac{1+\left\{f\right(x){\}}^2}{1+f\left(x\right)}+\frac{1}{f\left(x\right)+\sqrt{f\left(x\right)}+1}]\mathrm{dx}$

$=\int [\frac{1+x^4}{1+x^2}+\frac{1}{x^2+x+1}]\mathrm{dx}$

$=\int \frac{x^2(1+x^2)+1-x^2}{1+x^2}dx+\int \frac{1}{x^2+2\cdot x\cdot \frac{1}{2}+\frac{1}{4}+1-\frac{1}{4}}dx$

$=\int x^{2}dx+\int \frac{1-x^2}{1+x^2}dx+\int \frac{1}{{(x+\frac{1}{2})}^2+\frac{3}{4}}dx$

$=\frac{x^3}{3}+\int \frac{-(1+x^2)+2}{1+x^2}dx+\int \frac{1}{{(x+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}dx$

$=\frac{x^3}{3}-\int 1dx+2\int \frac{1}{1+x^2}dx+\frac{1}{\frac{\sqrt{3}}{2}}{\tan }^{-1}\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}$

$=\frac{x^3}{3}-x+2{\tan }^{-1}x+\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+c$

নির্ণেয় যোজিত ফল : $\frac{x^3}{3}-x+2{\tan }^{-1}x+\frac{2}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x+1}{\sqrt{3}}\right)+c$.

দেওয়া আছে, $f\left(x\right)=x^2$

$\begin{array}{cc}\mid & \therefore {\int }_1^2[\sqrt{f\left(x\right)}{e}^{f\left(x\right)}+\frac{f\left(x\right)-1{\}}^2}{f\left(x\right)}]dx={\int }_1^2[\sqrt{x^2}{e}^{x^2}+\frac{{(x^2-1)}^2}{x^2}]dx\\ & ={\int }_1^{2}x{e}^{x^2}dx+{\int }_1^2\frac{x^4-2x^2+1}{x^2}dx\\ & =\frac{1}{2}{\int }_1^{2}2x{e}^{x^2}dx+{\int }_1^2(x^2-2+x^{-2})dx\\ & =\frac{1}{2}{\int }_1^2{e}^{x^2}d\left(x^2\right)+{[\frac{x^3}{3}-2x+\frac{x^{-2+1}}{-2+1}]}_1^2=\frac{1}{2}{\left[e^{x^2}\right]}_1^2+{[\frac{x^3}{3}-2x-\frac{1}{x}]}_1^2\\ & =\frac{1}{2}(e^4-e^1)+(\frac{8}{3}-4-\frac{1}{2})-(\frac{1}{3}-2-1)\\ & =\frac{1}{2}(e^4-e)+\frac{16-24-3}{6}-\frac{1-9}{3}\\ & =\frac{1}{2}(e^4-e)+\frac{-11}{6}+\frac{8}{3}=\frac{1}{2}(e^4-e)+\frac{8}{3}-\frac{11}{6}\\ & =\frac{1}{2}(e^4-e)+\frac{16-11}{6}\\ & =\frac{1}{2}(e^4-e)+\frac{5}{6}\end{array}$

নির্ণেয় মান : $\frac{1}{2}(e^4-e)+\frac{5}{6}$.