UV আকারের (Integration by parts)
∫x3ex2dx=?\int x^{3} e^{x^{2}} d x = ?∫x3ex2dx=?
−12(x2+1)ex2+c-\frac{1}{2}\left(x^{2}+1\right) e^{x^{2}}+c−21(x2+1)ex2+c
12(x2+1)ex2+c\frac{1}{2}\left(x^{2}+1\right) e^{x^{2}}+c21(x2+1)ex2+c
−12(x2−1)ex2+c-\frac{1}{2}\left(x^{2}-1\right) e^{x^{2}}+c−21(x2−1)ex2+c
12(x2−1)ex2+c\frac{1}{2}\left(x^{2}-1\right) e^{x^{2}}+c21(x2−1)ex2+c
Solve: ধরি, I=∫x3ex2dx \mathrm{I}=\int x^{3} e^{x^{2}} d x I=∫x3ex2dx এবং x2=z \mathrm{x}^{2}=\mathrm{z} x2=z. তাহলে
2xdx=dz⇒xdx=12dz এবং I=∫x2ex2(xdx)=12∫zezdz=12[z∫ezdz−∫{ddz(z)∫ezdz}dz]=12[zez−∫1⋅ezdz]=12(zez−ez)+c=12(x2−1)ex2+c \begin{array}{l} 2 x d x=d z \Rightarrow x d x=\frac{1}{2} d z \text { এবং } \\ \mathrm{I}=\int x^{2} e^{x^{2}}(x d x)=\frac{1}{2} \int z e^{z} d z \\ =\frac{1}{2}\left[z \int e^{z} d z-\int\left\{\frac{d}{d z}(z) \int e^{z} d z\right\} d z\right] \\ =\frac{1}{2}\left[z e^{z}-\int 1 \cdot e^{z} d z\right]=\frac{1}{2}\left(z e^{z}-e^{z}\right)+\mathrm{c} \\ =\frac{1}{2}\left(x^{2}-1\right) e^{x^{2}}+c \end{array} 2xdx=dz⇒xdx=21dz এবং I=∫x2ex2(xdx)=21∫zezdz=21[z∫ezdz−∫{dzd(z)∫ezdz}dz]=21[zez−∫1⋅ezdz]=21(zez−ez)+c=21(x2−1)ex2+c
∫xcos−1x2dx=? \int x \cos ^{-1} x^{2} d x = ?∫xcos−1x2dx=?
∫ln(1+x)1+xdx equals\displaystyle\int {\dfrac{{\ln \left( {1 + {x}} \right)}}{{1 + {x}}}} dx\,equals∫1+xln(1+x)dxequals
∫x3exdx=f(x)+c \int x^{3} e^{x} dx = f{\left ( x \right )} + c ∫x3exdx=f(x)+c হয় তবে f(x)=?
The sequence S0,S1,S2S_0,S_1,S_2S0,S1,S2.... forms a G.P with common ratio
