Tìm:

a) $\int \left(3\sqrt{x}+\frac{1}{\sqrt[3]{x}}\right)dx$

b) $\int \sqrt{x}\left(7x^2-3\right)dx\left(x>0\right)$

c) $\int \frac{{\left(2x+1\right)}^2}{x^2}dx$

d) $\int \left(2^x+\frac{3}{x^2}\right)dx$

Lời giải của Vua Trắc Nghiệm

a) Ta có \(\int {\left( {3\sqrt x + \frac{1}{{\sqrt[3]{x}}}} \right)} dx\)\( = 3\int {{x^{\frac{1}{2}}}dx + \int {{x^{ – \frac{1}{3}}}dx} } = 2{x^{\frac{3}{2}}} + \frac{3}{2}{x^{\frac{2}{3}}} + C\)\( = 2x\sqrt x + \frac{3}{2}\sqrt[3]{{{x^2}}} + C\).

b) Ta có \(\int {\sqrt x \left( {7{x^2} – 3} \right)dx} \)\( = 7\int {{x^{\frac{5}{2}}}dx – 3\int {{x^{\frac{1}{2}}}dx} } = 2{x^{\frac{7}{2}}} – 2{x^{\frac{3}{2}}} + C = 2\sqrt {{x^7}} – 2\sqrt {{x^3}} + C\).

c) Ta có \(\int {\frac{{{{\left( {2x + 1} \right)}^2}}}{{{x^2}}}dx} \)\( = \int {\frac{{4{x^2} + 4x + 1}}{{{x^2}}}dx} \)\( = \int {\left( {4 + \frac{4}{x} + \frac{1}{{{x^2}}}} \right)dx} \)

\( = 4\int {dx + 4\int {\frac{1}{x}dx + \int {\frac{1}{{{x^2}}}} } } dx\)\( = 4x + 4\ln \left| x \right| – \frac{1}{x} + C\).

d) Ta có \(\int {\left( {{2^x} + \frac{3}{{{x^2}}}} \right)} dx\)\( = \int {{2^x}dx + 3\int {{x^{ – 2}}dx} } \)\( = \frac{{{2^x}}}{{\ln 2}} – \frac{3}{x} + C\).