1/ \(\left\{{}\begin{matrix}u=ln\left(x+1\right)\\dv=dx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=\dfrac{1}{x+1}dx\\v=x\end{matrix}\right.\)
\(\Rightarrow I=\int ln\left(1+x\right)dx=x.ln\left(x+1\right)-\int\dfrac{x}{x+1}dx\)
Xet \(I_1=\int\dfrac{x}{x+1}dx=\int(1-\dfrac{1}{x+1})dx=\int dx-\int\dfrac{dx}{x+1}=x-ln\left|x+1\right|\)
\(\Rightarrow I=x.ln\left(x+1\right)-x+ln\left|x+1\right|\)
2/ \(\left\{{}\begin{matrix}u=x^2-2x\\dv=e^{2x+1}dx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}du=\left(2x-2\right)dx\\v=\dfrac{1}{2}.e^{2x+1}\end{matrix}\right.\)
\(\Rightarrow I=\int\left(x^2-2x\right)e^{2x+1}dx=\dfrac{1}{2}\left(x^2-2x\right).e^{2x+1}-\int\left(x-1\right).e^{2x+1}dx\)
\(I_1=\int\left(x-1\right)e^{2x+1}dx\)
\(\left\{{}\begin{matrix}t=x-1\\dy=e^{2x+1}dx\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}dt=dx\\y=\dfrac{1}{2}.e^{2x+1}\end{matrix}\right.\)
\(\Rightarrow I_1=\dfrac{1}{2}\left(x-1\right)e^{2x+1}-\dfrac{1}{2}\int e^{2x+1}dx=\dfrac{1}{2}\left(x-1\right)e^{2x+1}-\dfrac{1}{4}e^{2x+1}\)
\(\Rightarrow I=\dfrac{1}{2}\left(x^2-2x\right)e^{2x+1}-\dfrac{1}{2}\left(x-1\right)e^{2x+1}+\dfrac{1}{4}e^{2x+1}\)
P/s: Bạn tự thay cận vô nhé!
