প্রতিস্থাপন পদ্ধতি (Method of Substitution)
∫1x1+lnx dx\int{\frac{1}{x\sqrt{1+lnx\ }}dx}∫x1+lnx 1dx এর মান হল-
121+lnx+c\frac{1}{2\sqrt{1+lnx}}+c21+lnx1+c
tan−1(1+lnx)+c\tan^{-1}{\left(\sqrt{1+lnx}\right)+c}tan−1(1+lnx)+c
2 1+lnx+c2\ \sqrt{1+lnx}+c2 1+lnx+c
ln(1+lnx)+c\ln{\left(\sqrt{1+lnx}\right)}+cln(1+lnx)+c
∫1x1+lnx dx[1+lnx=z ⇒1x.dx=dz]\int{\frac{1}{x\sqrt{1+lnx}}\ dx} [1+lnx=z\ \Rightarrow\frac{1}{x.}dx=dz] ∫x1+lnx1 dx[1+lnx=z ⇒x.1dx=dz]
=∫dzz=2z+c=2 1+lnx+c=\int{\frac{dz}{\sqrt z}=2\sqrt z+c=2\ \sqrt{1+lnx}+c}=∫zdz=2z+c=2 1+lnx+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?
