Đề bài

Tìm a) \(\int {\left( {2{x^5} + 3} \right)dx} \) b) \(\int {\left( {5\cos x - 3\sin x} \right)dx} \) c) \(\int {\left( {\frac{{\sqrt x }}{2} - \frac{2}{x}} \right)dx} \) d) \(\int {\left( {{e^{x - 2}} - \frac{2}{{{{\sin }^2}x}}} \right)dx} \)

Lời giải chi tiết

a) \(\int {\left( {2{x^5} + 3} \right)dx}  = 2\int {{x^5}dx}  + 3\int {dx}  = 2\frac{{{x^6}}}{6} + 3x + C = \frac{{{x^6}}}{3} + 3x + C\) b) \(\int {\left( {5\cos x - 3\sin x} \right)dx}  = 5\int {\cos xdx}  - 3\int {\sin xdx}  = 5\sin x - 3\left( { - \cos x} \right) + C\) \( = 5\sin x + 3\cos x + C\) c) \(\int {\left( {\frac{{\sqrt x }}{2} - \frac{2}{x}} \right)dx}  = \frac{1}{2}\int {{x^{\frac{1}{2}}}dx}  - 2\int {\frac{1}{x}dx}  = \frac{1}{2}.\frac{{{x^{\frac{3}{2}}}}}{{\frac{3}{2}}} - 2\ln \left| x \right| + C = \frac{1}{3}\sqrt {{x^3}}  - 2\ln \left| x \right| + C\) d) \(\int {\left( {{e^{x - 2}} - \frac{2}{{{{\sin }^2}x}}} \right)dx}  = {e^{ - 2}}\int {{e^x}dx}  - 2\int {\frac{1}{{{{\sin }^2}x}}dx = {e^{ - 2}}.{e^x} - 2\left( { - \cot x} \right) + C} \) \( = {e^{x - 2}} + 2\cot x + C\)