Ta có: F'(u) = f(u), với mọi u ∈ K.

\({\left[ {F\left( {u\left( x \right)} \right)} \right]^\prime }\) = \(F'\left( {u\left( x \right)} \right)\).u'(x) = \(f\left( {u\left( x \right)} \right).u'\left( x \right)\), với mọi x ∈ J.

Do đó, \(\int {f\left( {u(x)} \right).u'(x)dx} \) = \(F\left( {u\left( x \right)} \right)\) + C.

Áp dụng:

\(\int {{{\left( {2x + 1} \right)}^5}dx} \) = \({\int {\left( {2x + 1} \right)} ^5}\frac{{{{\left( {2x + 1} \right)}^\prime }}}{2}dx\)

                      = \(\frac{1}{2}\int {{{\left( {2x + 1} \right)}^5}{{\left( {2x + 1} \right)}^\prime }dx} \)

                      = \(\frac{1}{2}.\frac{{{{\left( {2x + 1} \right)}^6}}}{6} + C\)

                      = \(\frac{{{{\left( {2x + 1} \right)}^6}}}{{12}} + C\).

 \(\int {\frac{1}{{\sqrt {2x + 1} }}dx} \) = \(\int {\frac{1}{{\sqrt {2x + 1} }}.\frac{{{{\left( {2x + 1} \right)}^\prime }}}{2}dx} \)

                       = \(\int {\frac{1}{{2\sqrt {2x + 1} }}.{{\left( {2x + 1} \right)}^\prime }dx} \)

                       = \(\sqrt {2x + 1} + C.\)