a/ Đặt \(lnx=t\Rightarrow\frac{dx}{x}=dt\)
\(\Rightarrow I=\int\frac{t^2+1}{2}dt=\int\left(\frac{1}{2}t^2+\frac{1}{2}\right)dt=\frac{t^3}{6}+\frac{t}{2}+C\)
\(=\frac{ln^3x}{6}+\frac{lnx}{2}+C\)
b/ \(I=\int sin^2x.cos^2x.cosxdx=\int sin^2x\left(1-sin^2x\right)cosxdx\)
Đặt \(sinx=t\Rightarrow cosxdx=dt\)
\(I=\int t^2\left(1-t^2\right)dt=\int\left(t^2-t^4\right)dt=\frac{t^3}{3}-\frac{t^5}{5}+C\)
\(=\frac{1}{3}sin^3x-\frac{1}{5}sin^5x+C\)
c/ \(I=\int x^4\sqrt{x^2+1}xdx\)
Đặt \(\sqrt{x^2+1}=t\Rightarrow x^2=t^2-1\Rightarrow xdx=tdt\)
\(\Rightarrow I=\int\left(t^2-1\right)^2.t.tdt=\int\left(t^4-2t^2+1\right)t^2dt\)
\(=\int\left(t^6-2t^4+t^2\right)dt=\frac{1}{7}t^7-\frac{2}{5}t^5+\frac{1}{3}t^3+C\)
\(=\frac{1}{7}\sqrt{\left(x^2+1\right)^7}-\frac{2}{5}\sqrt{\left(x^2+1\right)^5}+\frac{1}{3}\sqrt{\left(x^2+1\right)^3}+C\)
d/ Đặt \(1+\sqrt{x}=t\Rightarrow x=\left(t-1\right)^2\Rightarrow dx=2\left(t-1\right)dt\)
\(\Rightarrow I=\int\frac{2\left(t-1\right)dt}{t}=\int\left(2-\frac{2}{t}\right)dt=2t-2lnt+C\)
\(=2\left(1+\sqrt{x}\right)-2ln\left(1+\sqrt{x}\right)+C\)
