প্রতিস্থাপন পদ্ধতি (Method of Substitution)
∫dxex+e−x=?\int{\frac{dx}{e^x+e^{-x}}=?}∫ex+e−xdx=?
sin−1ex+c\sin^{-1}{e^x}+c sin−1ex+c
tan−1ex+c \tan^{-1}{e^x}+c tan−1ex+c
tan−11ex+ctan^{-1}{\frac{1}{e^x}}+c tan−1ex1+c
cos−1ex+c \cos^{-1}{e^x}+ccos−1ex+c
∫dxex+e−x=∫exdx(ex)2+1=∫dzz2+1\int{\frac{dx}{e^x+e^{-x}}=\int{\frac{e^xdx}{\left(e^x\right)^2+1}=\int\frac{dz}{z^2+1}}}∫ex+e−xdx=∫(ex)2+1exdx=∫z2+1dz [ধরি, z=ex]= tan−1(z)=tan−1(ex)+cz = e^x] =\ \tan^{-1}{(}z)=\tan^{-1}{(}e^x)+cz=ex]= tan−1(z)=tan−1(ex)+c
∫ex(x+1)dxsin2(xex)=? \int \frac{e^{x} \left ( x + 1 \right ) dx}{\sin^{2}{\left ( x e^{x} \right )}} = ? ∫sin2(xex)ex(x+1)dx=?
∫exdx1+e2x=f(x)+c \int \frac{e^{x} dx}{1 + e^{2 x}} = f{\left ( x \right )} + c ∫1+e2xexdx=f(x)+c
হলে, f(x)=?
∫tan(sin−1x)1−x2dx=?\int_{ }^{ }\frac{\tan\left(\sin^{-1}x\right)}{\sqrt{1-x^2}}dx=?∫1−x2tan(sin−1x)dx=?
What is ∫x4−1x2x4+x2+1dx\displaystyle \int \dfrac{x^4 - 1}{x^2 \sqrt{x^4 + x^2 + 1}} dx∫x2x4+x2+1x4−1dx equal to?
