প্রতিস্থাপন পদ্ধতি (Method of Substitution)
∫dx(ex−1)2=?\int \frac{d x}{\left(e^{x}-1\right)^{2}} = ?∫(ex−1)2dx=?
ln∣1−e−x∣+11−e−x+c\ln \left|1-e^{-x}\right|+\frac{1}{1-e^{-x}}+cln∣1−e−x∣+1−e−x1+c
−ln∣1+e−x∣−11−e−x+c-\ln \left|1+e^{-x}\right|-\frac{1}{1-e^{-x}}+c−ln∣1+e−x∣−1−e−x1+c
−ln∣1−e−x∣−11−e−x+c-\ln \left|1-e^{-x}\right|-\frac{1}{1-e^{-x}}+c−ln∣1−e−x∣−1−e−x1+c
ln∣1−e−x∣−11−e−x+c\ln \left|1-e^{-x}\right|-\frac{1}{1-e^{-x}}+cln∣1−e−x∣−1−e−x1+c
Solve: Let, I =∫dx(ex−1)2=∫dx{eI(1−e−x)}2 \text { Let, I }=\int \frac{d x}{\left(e^{x}-1\right)^{2}}=\int \frac{d x}{\left\{e^{\mathbb{I}}\left(1-e^{-x}\right)\right\}^{2}} Let, I =∫(ex−1)2dx=∫{eI(1−e−x)}2dx
=∫dxe2x(1−e−x)2=∫e−x⋅e−xdx(1−e−x)2 এবং e−x=z⋅ Then −e−xdx=dz and I=−∫zdz(1−z)2=∫(1−z)−1(1−z)2dz=∫{11−z−1(1−z)2}dz=−∫{11−z−1(1−z)2}d(1−z)=−{ln∣1−z∣+11−z}+c=−ln∣1−e−x∣−11−e−x+c \begin{array}{l} =\int \frac{d x}{e^{2 x}\left(1-e^{-x}\right)^{2}}=\int \frac{e^{-x} \cdot e^{-x} d x}{\left(1-e^{-x}\right)^{2}} \text { এবং } \\ e^{-x}=z \cdot \text { Then }-e^{-x} d x=d z \text { and } \\ \mathrm{I}=-\int \frac{z d z}{(1-z)^{2}}=\int \frac{(1-z)-1}{(1-z)^{2}} d z \\ =\int\left\{\frac{1}{1-z}-\frac{1}{(1-z)^{2}}\right\} d z \\ =-\int\left\{\frac{1}{1-z}-\frac{1}{(1-z)^{2}}\right\} d(1-z) \\ =-\left\{\ln |1-z|+\frac{1}{1-z}\right\}+c \\ =-\ln \left|1-e^{-x}\right|-\frac{1}{1-e^{-x}}+c \end{array} =∫e2x(1−e−x)2dx=∫(1−e−x)2e−x⋅e−xdx এবং e−x=z⋅ Then −e−xdx=dz and I=−∫(1−z)2zdz=∫(1−z)2(1−z)−1dz=∫{1−z1−(1−z)21}dz=−∫{1−z1−(1−z)21}d(1−z)=−{ln∣1−z∣+1−z1}+c=−ln∣1−e−x∣−1−e−x1+c
∫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)=?
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?
∫ecos−1x1−x2dx \int \frac{e^{\cos^{- 1}{x}}}{\sqrt{1 - x ²}} dx ∫1−x2ecos−1xdx এর মান কত?
∫cosxdx(1−sinx)2=? \int \frac{\cos x d x}{(1-\sin x)^{2}} \quad = ?∫(1−sinx)2cosxdx=?
