সৃজনশীলঃ ‘…’ আইডিয়াল স্কুল অ্যান্ড কলেজ ২০২৪ – যোগজীকরণ, উচ্চতর গণিত, উচ্চতর গণিত ১ম পত্র (এইচএসসি)

(i)

$\int \frac{dx}{(2x+1)\sqrt{2x+5}}$

$=\int \frac{dx}{(2x+5-4)\sqrt{2x+5}}$

$=\int \frac{dxzdz}{(z^2-4)\sqrt{z^2}}$

let,

$z^2=2x+5$

$2z\cdot \frac{dz}{dx}=2$

$\Rightarrow zdz=dx$

$=\int \frac{zdz}{(z^2-4)z}$

$=\int \frac{dz}{z^2-2^2}$

$=\frac{1}{2\cdot 2}\ln \left|\frac{z-2}{z+2}\right|+c$

$=\frac{1}{4}\ln \left|\frac{2x+5-2}{2x+5+2}\right|+c$

$=\frac{1}{4}\ln \left|\frac{2x+3}{2x+7}\right|+C\text{ Ans. }$

(ii)

$\int \frac{x^2}{x^2-4}dx$

= $\int \frac{x^2-4+4}{x^2-4}dx$

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

= $\int dx+\int \frac{4dx}{x^2-4}$

= $\int dx+4\int \frac{dx}{x^2-2^2}$

= $x+4\cdot \frac{1}{2\cdot 2}\ln \left|\frac{x-2}{x+2}\right|+c$

= $x+\ln \left|\frac{x-2}{x+2}\right|+c\text{ Ans. }$

উদ্দীপক হতে পাই,

$x^2=4y$

$y=\frac{x^2}{4}$

$y^2=4x$

$\Rightarrow {\left(\frac{x^2}{4}\right)}^2=4x$ [(i) নং হতে]

$\Rightarrow \frac{x^4}{16}-4x=0$

$\Rightarrow x^4-64x=0$

$\Rightarrow x(x^3-64)=0$

হয়, x = 0

অথবা,

$x^3-64=0$

$\Rightarrow x^3=64$

$\therefore x=4$

$\therefore$ ছায়াঘেরা অংশের ক্ষেত্রফল = ${\int }_0^4(2\sqrt{x}-\frac{x^2}{4})dx$

$={\int }_0^4(2\cdot x^{1/2}-\frac{x^2}{4})dx$

$={[\frac{2x^{3/2}}{3/2}-\frac{1}{4}\cdot \frac{x^3}{3}]}_0^4$

$={[\frac{4}{3}{x}^{3/2}-\frac{1}{12}{x}^3]}_0^4$

$=[(\frac{4}{3}{4}^{3/2}-\frac{1}{12}{4}^3)-(\frac{4}{3}{0}^{3/2}-\frac{1}{12}{0}^3)]$

$=\frac{4}{3}.8-\frac{64}{12}$

$=\frac{32}{3}-\frac{16}{3}$

$=\frac{32-16}{3}.$

$=\frac{16}{3}\text{ Ars. }$