$\int \sqrt{e^{x}} d x=\int\left(e^{x}\right)^{\frac{1}{2}}=\int e^{\frac{x}{2}}=2 e^{\frac{x}{2}}+c$

$\int \frac{\tan \left(\sin ^{-1} x\right)}{\sqrt{1-x^{2}}}$ ধরি, $\sin ^{-1} x=z$
$\int \tan z=\ln \sec z+c \quad$
$=\ln \left[\sec \left(\sin ^{-1} x\right)\right]+c$

$\int \frac{\mathrm{e}^{\mathrm{x}}-\mathrm{e}^{-\mathrm{x}}}{\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}} \mathrm{dx}\left[\because \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\mathrm{f}(\mathrm{x})} \mathrm{dx}=\ln \{\mathrm{f}(\mathrm{x})\}+\mathrm{c}\right]$
$=\ln \left(e^{x}+e^{-x}\right)+c$
$\therefore \int \frac{\mathrm{e}^{\mathrm{x}}-\mathrm{e}^{-\mathrm{x}}}{\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}} \mathrm{dx}=\mathrm{f}(\mathrm{x})+\mathrm{c} \quad$
$\therefore \mathrm{f}(\mathrm{x})=\ln \left(\mathrm{e}^{\mathrm{x}}+\mathrm{e}^{-\mathrm{x}}\right)$

$\int \frac{1}{3 \sqrt{x}} d x=\frac{2}{3} \int \frac{1}{2 \sqrt{x}} d x=\frac{2}{3} \sqrt{x}+c$

$\int \frac{d x}{9+4 x^{2}}=\int \frac{d x}{3^{2}+(2 x)^{2}}$
$=\frac{1}{3} \cdot \tan ^{-1} \frac{2 x}{3} \cdot \frac{1}{2}+\mathrm{c}=\frac{1}{6} \tan ^{-1} \frac{2 x}{3}+\mathrm{c}$

$\int \frac{d x}{\sqrt{36-x^{2}}}=\int \frac{d x}{\sqrt{6^{2}-x^{2}}}=\sin ^{-1} \frac{x}{6}+c$

$\int \frac{d x}{1+\cos 2 x}=\int \frac{d x}{2 \cos ^{2} x}$
$=\frac{1}{2} \int \sec ^{2} x d x=\frac{1}{2} \tan x+c$

$\int \frac{1}{x(1+\ln (x))} d x=\int \frac{\frac{1}{x} d x}{1+\ln x}$
$\left[\because \int \frac{f^{\prime}(x)}{f(x)} d x=\ln \{\mathrm{f}(\mathrm{x})\}+\mathrm{c}\right]$
$=\ln (1+\ln x)+c$

$\int_{0}^{\pi / 2} \sin ^{2} x d x=\frac{1}{2} \int_{0}^{\pi / 2}(1-\cos 2 x) d x$
$=\frac{1}{2}\left[x-\frac{1}{2} \sin 2 x\right]_{0}^{\pi / 2}=\frac{1}{2}\left(\frac{\pi}{2}-\frac{1}{2} \sin \pi\right)$
$=\pi / 4$

$\int_{0}^{1} \frac{d x}{\sqrt{2 x-x^{2}}}=\int_{0}^{1} \frac{d x}{\sqrt{1-1+2 x-x^{2}}}$
$=\int_{0}^{1} \frac{d x}{\sqrt{1-(x-1)^{2}}}=\left[\sin ^{-1}(x-1)\right]_{0}^{1}$
$=\left[\sin ^{-1} 0-\sin ^{-1}(-1)\right]=\frac{\pi}{2}$

ধরি, $\ln (x+1)=z$ $\therefore \frac{1}{\mathrm{x}+1} \mathrm{dx}=\mathrm{dz}$
$\therefore \int \frac{\ln (x+1)}{x+1} \mathrm{dx}=\int \mathrm{zdz}=\frac{\mathrm{z}^{2}}{2}$
$=\frac{\{\ln (\mathrm{x}+1)\}^{2}}{2}=\int_{0}^{1} \frac{\ln (\mathrm{x}+1)}{\mathrm{x}+1} \mathrm{dx}$
$=\left[\frac{\{\ln (x+1)\}^{2}}{2}\right]_{0}^{1}=\frac{1}{2}(\ln 2)^{2}$

$\int_{0}^{4} \sqrt{3 x+4} \mathrm{~d} x=\left[\frac{(3 x+4)^{3 / 2}}{3 / 2} \cdot \frac{1}{3}\right]_{0}^{4}$
$=\frac{2}{9}\left[(3 x+4)^{3 / 2}\right]^{4}=\frac{2}{9}\left[(16)^{3 / 2}-(4)^{3 / 2}\right]$
$=\frac{2}{9}(64-8)=\frac{2}{9} \times 56=\frac{112}{9}$

ক্ষেত্রফল$=\frac{8 a^{2}}{3 m^{3}}=\frac{8 \times(4)^{2}}{3(4)^{3}}=\frac{2}{3}$ বর্গ একক

ক্ষেত্রফল$=\frac{8}{3} a^{2} m^{3}=\frac{8}{3} \times\left(\frac{1}{4}\right)^{2} \times(1)^{3}=\frac{1}{6}$বর্গ একক