প্রতিস্থাপন পদ্ধতি (Method of Substitution)
∫dx(x−4)(5−x)=?\int{\frac{dx}{\sqrt{\left(x-4\right)\left(5-x\right)}}=?} ∫(x−4)(5−x)dx=?
2sin−15−x+c2\sin^{-1}{\sqrt{5-x}+c} 2sin−15−x+c
2sin−1x−4+c2\sin^{-1}{\sqrt{x-4}}+c 2sin−1x−4+c
sin−15−x+c\sin^{-1}{\sqrt{5-x}+c} sin−15−x+c
2ln∣x−4+5−x∣+c 2\ln{\left|\sqrt{x-4}+\sqrt{5-x}\right|+c} 2lnx−4+5−x+c
ধরি, x−4=z2⇒dx=2zdz∴x−4=z2⇒−x+4+1=−z2+1⇒−x+5=1−z2 \begin{array}{l}\text { ধরি, } x-4=z^{2} \\ \quad \Rightarrow d x=2 z d z \\ \therefore x-4=z^{2} \\ \Rightarrow-x+4+1=-z^{2}+1 \\ \Rightarrow-x+5=1-z^{2}\end{array} ধরি, x−4=z2⇒dx=2zdz∴x−4=z2⇒−x+4+1=−z2+1⇒−x+5=1−z2
∫dx(x−4)(5−x)=∫2zdzz2(1−z2)=∫2zdzz⋅1−z2=∫dz1−z2 \begin{aligned} & \int \frac{d x}{\sqrt{(x-4)(5-x)}} \\ = & \int \frac{2 z d z}{\sqrt{z^{2}\left(1-z^{2}\right)}} \\ = & \int \frac{2 z d z}{z \cdot \sqrt{1-z^{2}}} \\ = & \int \frac{d z}{\sqrt{1-z^{2}}}\end{aligned} ===∫(x−4)(5−x)dx∫z2(1−z2)2zdz∫z⋅1−z22zdz∫1−z2dz
[∵∫dx1−x2=sin−1x] \left[\because \int \frac{d x}{\sqrt{1-x^{2}}}=\sin ^{-1} x\right] [∵∫1−x2dx=sin−1x]
=2sin−1z+c=2sin−1x−4+c \begin{array}{l}=2 \sin ^{-1} z+c \\ =2 \sin ^{-1} \sqrt{x-4}+c\end{array} =2sin−1z+c=2sin−1x−4+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?
