1,
đặt $t=\sqrt{x^{2}+1}\Rightarrow tdt=xdx$
$I=\int\limits_{1}^{2}(t^{4}-1)dt=(\frac{t^{5}}{5}-t)|^{2}_{1}=\frac{26}{5}$
2,
đặt $t=\sqrt[3]{2x+2}\Rightarrow dx=\frac{3t^{2}}{2}$ và $x=\frac{t^{3}-2}{2}$
$I=\frac{3}{4}\int\limits_{1}^{2}(t^{4}-2t)dt=\frac{3}{4}(\frac{t^{5}}{5}-t^{2})|^{2}_{1}=\frac{12}{5}$
3.
$I=\int\limits_{0}^{1}\frac{dx}{x^{2}+1}+\frac{1}{3}\int\limits_{0}^{1}\frac{d(x^{3})}{1+(x^{3})^{2}}=I_{1}+I_{2}$
đặt $x=tant\Rightarrow dx=(1+tan^{2}t)dt$
$I_{1}=\int\limits_{0}^{\frac{\pi }{4}}tdt=\frac{\pi }{4}$
đặt $x^{3}=tant\Rightarrow dx^{3}=(1+tan^{2}t)dt$
$I_{2}=\frac{1}{3}\int\limits_{0}^{\frac{\pi }{4}}dt=\frac{\pi }{12}$
do đó $I=\frac{\pi }{3}$
5,
$I=\int\limits_{0}^{\frac{\pi }{2}}e^{sinx}dsinx+\int\limits_{0}^{\frac{\pi }{2}}cos^{2}xdx$
$=e^{sinx}|^{\frac{\pi }{2}}_{0}+\frac{1}{2}\int\limits_{0}^{\frac{\pi }{2}}(1+cos2x)dx=e-1+\frac{1}{2}(x+\frac{1}{2}sin2x)|^{\frac{\pi }{2}}_{0}=e+\frac{\pi }{4}-1$
6,
$I=\int\limits_{0}^{1}\frac{(x^{2}+3x+1)e^{x}+1}{(x+1)(xe^{x}+1)}dx-2\int\limits_{0}^{1}\frac{dx}{x+1}$
$=ln|(x+1)(xe^{x}+1)||^{1}_{0}-2ln|x+1||^{1}_{0}=ln(e+1)-ln2.$