যোগজীকরণ এর অন্যান্য
∫5e2x1+e4xdx+? \int \frac{5 e^{2x}}{1+e^{4x}}dx+?∫1+e4x5e2xdx+?
12tan−1(e5x)+c\frac{1}{2} \tan ^{-1}\left(\mathrm{e}^{5 \mathrm{x}}\right)+\mathrm{c}21tan−1(e5x)+c
12tan−1(e2x)+c\frac{1}{2} \tan ^{-1}\left(\mathrm{e}^{2 \mathrm{x}}\right)+\mathrm{c}21tan−1(e2x)+c
52tan−1(e2x)+c\frac{5}{2} \tan ^{-1}\left(\mathrm{e}^{2 \mathrm{x}}\right)+\mathrm{c}25tan−1(e2x)+c
15tan−1(e4x)+c\frac{1}{5} \tan ^{-1}\left(\mathrm{e}^{4 \mathrm{x}}\right)+\mathrm{c}51tan−1(e4x)+c
e2x=z∴e2xdx=dz2∫5e2x1+(e2x)2dx=5∫(dz)/21+z2=52∫dz1+z2=52tan−1(e2x)+c \mathrm{e}^{2 \mathrm{x}}=\mathrm{z} \quad \therefore \mathrm{e}^{2 \mathrm{x}} \mathrm{dx}=\frac{\mathrm{dz}}{2} \int \frac{5 \mathrm{e}^{2 \mathrm{x}}}{1+\left(\mathrm{e}^{2 \mathrm{x}}\right)^{2}} \mathrm{dx}=5 \int \frac{(\mathrm{dz}) / 2}{1+\mathrm{z}^{2}}=\frac{5}{2} \int \frac{\mathrm{dz}}{1+\mathrm{z}^{2}}=\frac{5}{2} \tan ^{-1}\left(\mathrm{e}^{2 \mathrm{x}}\right)+\mathrm{c} e2x=z∴e2xdx=2dz∫1+(e2x)25e2xdx=5∫1+z2(dz)/2=25∫1+z2dz=25tan−1(e2x)+c
∫aaax.aax.axdx=? \int a^{^{a^{a^{x}}}} . a^{a^{x}} . a^{x} dx = ? ∫aaax.aax.axdx=?
limx→0∫0x(tan−1t)21+x2dt \lim _{x \rightarrow 0} \int_{0}^{x} \frac{\left(\tan ^{-1} t\right)^{2}}{\sqrt{1+x^{2}}} d t limx→0∫0x1+x2(tan−1t)2dt is equal to-
Suppose a pyramid has a square base of length 3 m in each side and a height of 5 m. What is the volume of that pyramid?
