প্রতিস্থাপন পদ্ধতি (Method of Substitution)
∫xdx4−x=?\int \frac{x d x}{4-x} = ?∫4−xxdx=?
x−4ln∣4−x∣+cx-4 \ln |4-x|+cx−4ln∣4−x∣+c
−x−4ln∣4−x∣+c-x-4 \ln |4-x|+c−x−4ln∣4−x∣+c
x+4ln∣4−x∣+cx+4 \ln |4-x|+cx+4ln∣4−x∣+c
x−ln∣4−x∣+cx- \ln |4-x|+cx−ln∣4−x∣+c
Solve:
∫xdx4−x=∫−(4−x)+44−xdx=∫−(4−x)4−xdx+∫44−xdx=−∫dx−4∫d(4−x)4−x \begin{array}{l} \int \frac{x d x}{4-x}=\int \frac{-(4-x)+4}{4-x} d x \\ =\int \frac{-(4-x)}{4-x} d x+\int \frac{4}{4-x} d x \\ =-\int d x-4 \int \frac{d(4-x)}{4-x} \end{array} ∫4−xxdx=∫4−x−(4−x)+4dx=∫4−x−(4−x)dx+∫4−x4dx=−∫dx−4∫4−xd(4−x)
=x−4ln∣4−x∣+c =x-4 \ln |4-x|+c =x−4ln∣4−x∣+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=?
