a) Đặt \(u = \sqrt {x + 1} \Rightarrow {u^2} = x + 1 \Rightarrow 2udu = dx.\)
Đổi cận
\(\int\limits_0^1 {\sqrt {x + 1} } dx = \int\limits_1^{\sqrt 2 } {u.2udu = 2\int\limits_1^{\sqrt 2 } {{u^2}du} } = \left. {2.{{{u^3}} \over 3}} \right|_1^{\sqrt 2 } = {2 \over 3}\left( {2\sqrt 2 - 1} \right)\)
b) Đặt \(u = \tan x \Rightarrow du = {{dx} \over {{{\cos }^2}x}}\)
\(\int\limits_0^{{\pi \over 4}} {{{\tan x} \over {{{\cos }^2}x}}} dx = \int\limits_0^1 {udu = } \left. {{{{u^2}} \over 2}} \right|_0^1 = {1 \over 2}\)
c) Đặt \(u = 1 + {t^4} \Rightarrow du = 4{t^3}dt \Rightarrow {t^3}dt = {{du} \over 4}\)
\(\int\limits_0^1 {{t^3}\left( {1 + {t^4}} \right)} dt = {1 \over 4}\int\limits_1^2 {{u^3}} du = \left. {{1 \over 4}{{{u^4}} \over 4}} \right|_1^2 = {1 \over {16}}\left( {16 - 1} \right) = {{15} \over {16}}\)
d) Đặt \(u = {x^2} + 4 \Rightarrow du = 2xdx \Rightarrow xdx = {1 \over 2}du\)
\(\int\limits_0^1 {{{5x} \over {{{\left( {{x^2} + 4} \right)}^2}}}} dx = {5 \over 2}\int\limits_4^5 {{{du} \over {{u^2}}}} = \left. {{5 \over 2}\left( { - {1 \over u}} \right)} \right|_4^5 = {1 \over 8}\)
e) Đặt \(u = \sqrt {{x^2} + 1} \Rightarrow {u^2} = {x^2} + 1 \Rightarrow udu = xdx\)
\(\int\limits_0^{\sqrt 3 } {{{4x} \over {\sqrt {{x^2} + 1} }}} dx = 4\int\limits_1^2 {{{udu} \over u}} = \left. {4u} \right|_1^2 = 4\)
f) Đặt \(u = 1 - \cos 3x \Rightarrow du = 3\sin 3xdx \Rightarrow \sin 3xdx = {1 \over 3}du\)
\(\int\limits_0^{{\pi \over 6}} {\left( {1 - \cos 3x} \right)} \sin 3xdx = {1 \over 3}\int\limits_0^1 {udu = \left. {{{{u^2}} \over 6}} \right|} _0^1 = {1 \over 6}\)
