Tính $I = \int {\ln \left( {x + \sqrt {{x^2} + 1} } \right)dx}$ ta được:

$a \in \left( {\dfrac{1}{2};1} \right).$

$a \in \left( {1;\dfrac{3}{2}} \right).$

$a \in \left( {\dfrac{3}{2};2} \right).$

$a \in \left( {2;\dfrac{5}{2}} \right).$

$y=\tan x$                             

$y=\cot x$                             

$y=\sin x$                              

$F'\left( x \right) = f''\left( x \right)$

$F'\left( x \right) = f'\left( x \right)$             

$F'\left( x \right) = f\left( x \right)$

$f'\left( x \right) = F\left( x \right)$

$I = 12$

$I = 48$

$I = 8$

$I = 3$

$I = \left. {\left( {\dfrac{{{u^3}}}{3} + u} \right)} \right|_1^2$                  

$I = \left. {\dfrac{4}{3}\left( {{u^3} + u} \right)} \right|_1^2$

$I = \left. {2\left( {\dfrac{{{u^3}}}{3} + u} \right)} \right|_1^2$

$I = \left. {\dfrac{1}{3}\left( {\dfrac{{{u^3}}}{3} + u} \right)} \right|_1^2$

$\int\limits_1^{{e^2}} {\ln xdx}$.

$\int\limits_0^1 {2dx}$.

$\int\limits_0^\pi  {\sin xdx}$.

$\int\limits_0^2 {xdx}$.

$\int\limits_a^b {f'\left( x \right){e^{f\left( x \right)}}dx}  = 0$      

$\int\limits_a^b {f'\left( x \right){e^{f\left( x \right)}}dx}  = 1$

$\int\limits_a^b {f'\left( x \right){e^{f\left( x \right)}}dx}  =  - 1$  

$\int\limits_a^b {f'\left( x \right){e^{f\left( x \right)}}dx}  = 2$