UV আকারের (Integration by parts)
∫01lnxdx \int_{0}^{1} \ln x d x ∫01lnxdx
222
−2-2−2
−1-1−1
111
Solve:
∴∫lnxdx=lnx∫dx−∫{ddx(lnx)∫dx}dx=xlnx−∫1xxdx=xlnx−∫dx=xlnx−x+c=x(lnx−1)+c∴∫01lnxdx=[x(lnx−1)]01=(1.ln1−1)−0=−1( Ans: ) \begin{aligned} \therefore \quad & \int \ln x d x \\ = & \ln x \int d x-\int\left\{\frac{d}{d x}(\ln x) \int d x\right\} d x \\ = & x \ln x-\int \frac{1}{x} x d x=x \ln x-\int d x \\ = & x \ln x-x+c=x(\ln x-1)+c \\ \therefore \quad & \int_{0}^{1} \ln x d x=[x(\ln x-1)]_{0}^{1} \\ & =(1 . \ln 1-1)-0=-1(\text { Ans: }) \end{aligned} ∴===∴∫lnxdxlnx∫dx−∫{dxd(lnx)∫dx}dxxlnx−∫x1xdx=xlnx−∫dxxlnx−x+c=x(lnx−1)+c∫01lnxdx=[x(lnx−1)]01=(1.ln1−1)−0=−1( Ans: )
∫xcos−1x2dx=? \int x \cos ^{-1} x^{2} d x = ?∫xcos−1x2dx=?
∫ln(1+x)1+xdx equals\displaystyle\int {\dfrac{{\ln \left( {1 + {x}} \right)}}{{1 + {x}}}} dx\,equals∫1+xln(1+x)dxequals
∫x3exdx=f(x)+c \int x^{3} e^{x} dx = f{\left ( x \right )} + c ∫x3exdx=f(x)+c হয় তবে f(x)=?
The sequence S0,S1,S2S_0,S_1,S_2S0,S1,S2.... forms a G.P with common ratio
