a/ \(I=\int\limits^1_0\frac{x^4}{x^2+1}xdx\)
Đặt \(x^2+1=t\Rightarrow xdx=\frac{1}{2}dt\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow t=1\\x=1\Rightarrow t=2\end{matrix}\right.\)
\(\Rightarrow I=\int\limits^2_1\frac{\left(t-1\right)^2}{2t}dt=\int\limits^2_1\left(\frac{t}{2}-1+\frac{1}{2t}\right)dt=\left(\frac{1}{4}t^2-t+\frac{1}{2}ln\left|t\right|\right)|^2_1=\frac{1}{2}ln2-\frac{1}{4}\)
b/ \(I=\int\limits^1_0\frac{e^xdx}{e^x\left(e^x+1\right)}\)
Đặt \(e^x+1=t\Rightarrow e^xdx=dt\) ; \(\left\{{}\begin{matrix}x=0\Rightarrow t=2\\x=1\Rightarrow t=e+1\end{matrix}\right.\)
\(I=\int\limits^{e+1}_2\frac{dt}{t\left(t-1\right)}=\int\limits^{e+1}_2\left(\frac{1}{t-1}-\frac{1}{t}\right)dt=ln\left|\frac{t-1}{t}\right||^{e+1}_2=ln\left(\frac{e}{2e+2}\right)\)
Bài 2:
a/ \(I=\int\limits^{\frac{\pi}{2}}_0x.cosxdx+\int\limits^{\frac{\pi}{2}}_0sin^2x.cosxdx=I_1+I_2\)
Xét \(I_1=\int\limits^{\frac{\pi}{2}}_0x.cosxdx\)
Đặt \(\left\{{}\begin{matrix}u=x\\dv=cosxdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=dx\\v=sinx\end{matrix}\right.\)
\(\Rightarrow I_1=x.sinx|^{\frac{\pi}{2}}_0-\int\limits^{\frac{\pi}{2}}_0sinxdx=\left(xsinx+cosx\right)|^{\frac{\pi}{2}}_0=\frac{\pi}{2}-1\)
Xét \(I_2=\int\limits^{\frac{\pi}{2}}_0sin^2x.cosxdx=\int\limits^{\frac{\pi}{2}}_0sin^2x.d\left(sinx\right)=\frac{1}{3}sin^3x|^{\frac{\pi}{2}}_0=\frac{1}{3}\)
\(\Rightarrow I=\frac{\pi}{2}-1+\frac{1}{3}=\frac{\pi}{2}-\frac{2}{3}\)
b/ Đặt \(\left\{{}\begin{matrix}u=lnx\\dv=\frac{dx}{x^2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\frac{dx}{x}\\v=-\frac{1}{x}\end{matrix}\right.\)
\(\Rightarrow I=-\frac{lnx}{x}|^e_{\frac{1}{e}}+\int\limits^e_{\frac{1}{e}}\frac{dx}{x^2}=-e-\frac{1}{e}-\frac{1}{x}|^e_{\frac{1}{e}}=-\frac{2}{e}\)
