Question

Tìm:

a) \(\int\left(3x^2+\dfrac{2}{\sqrt[5]{x^3}}\right)dx\left(x>0\right);\)                   b) \(\int\left(\dfrac{3}{\cos^2x}-\dfrac{1}{\sin^2x}\right)dx.\)

Answer 1

a) Với \(x > 0\), ta có:

\(\int {\left( {3{x^3} + \frac{2}{{\sqrt[5]{{{x^3}}}}}} \right)dx}  = 3\int {{x^3}dx}  + 2\int {\frac{1}{{{x^{\frac{3}{5}}}}}dx = 3\int {{x^3}dx}  + 2\int {{x^{\frac{{ - 3}}{5}}}} dx = \frac{{3{x^4}}}{4} + \frac{{2{x^{\frac{2}{5}}}}}{{\frac{2}{5}}} + C} \)

\( = \frac{{3{x^4}}}{4} + 5\sqrt[5]{{{x^2}}} + C\)

b) \(\int {\left( {\frac{3}{{{{\cos }^2}x}} - \frac{1}{{{{\sin }^2}x}}} \right)dx}  = 3\int {\frac{1}{{{{\cos }^2}x}}dx - \int {\frac{1}{{{{\sin }^2}x}}dx}  = 3\tan x - \left( { - \cot x} \right) + C} \)

\( = 3\tan x + \cot x + C\)