Giải bài 23 trang 15 sách bài tập toán 12 - Cánh diều — Không quảng cáo

Đề bài

Tìm:

a) $\int {{x^{\frac{1}{3}}}} dx$;

b) $\int {\sqrt {\frac{1}{{{x^7}}}} } dx$;

c) $\int {\frac{1}{{\sqrt[3]{{{x^{\frac{4}{5}}}}}}}} dx$;

d) $\int {{{\left( {x - \frac{1}{x}} \right)}^2}} dx$;

e) $\int {\frac{{\left( {x - 3} \right)\left( {x + 1} \right)}}{x}} dx$;

g) $\int {\left( {3{{\rm{x}}^2} - \frac{4}{x}} \right)\left( {2{\rm{x}} + 5} \right)} dx$.

Lời giải chi tiết

a)

$\int {{x^{\frac{1}{3}}}} dx = \frac{{{x^{\frac{1}{3} + 1}}}}{{\frac{1}{3} + 1}} + C = \frac{{{x^{\frac{4}{3}}}}}{{\frac{4}{3}}} + C = \frac{3}{4}{x^{\frac{4}{3}}} + C$.

b)

$\int {\sqrt {\frac{1}{{{x^7}}}} } dx = \int {\frac{1}{{\sqrt {{x^7}} }}} dx = \int {\frac{1}{{{x^{\frac{7}{2}}}}}} dx = \int {{x^{ - \frac{7}{2}}}} dx = \frac{{{x^{ - \frac{7}{2} + 1}}}}{{ - \frac{7}{2} + 1}} + C = \frac{{{x^{ - \frac{5}{2}}}}}{{ - \frac{5}{2}}} + C =  - \frac{2}{5}{x^{ - \frac{5}{2}}} + C$.

c)

$\int {\frac{1}{{\sqrt[3]{{{x^{\frac{4}{5}}}}}}}} dx = \int {\frac{1}{{{x^{\frac{4}{{15}}}}}}} dx = \int {{x^{ - \frac{4}{{15}}}}} dx = \frac{{{x^{ - \frac{4}{{15}} + 1}}}}{{ - \frac{4}{{15}} + 1}} + C = \frac{{{x^{\frac{{11}}{{15}}}}}}{{\frac{{11}}{{15}}}} + C = \frac{{15}}{{11}}{x^{\frac{{11}}{{15}}}} + C$.

d)

$\begin{array}{l}\int {{{\left( {x - \frac{1}{x}} \right)}^2}} dx = \int {\left( {{x^2} - 2.x.\frac{1}{x} + \frac{1}{{{x^2}}}} \right)dx}  = \int {\left( {{x^2} - 2 + {x^{ - 2}}} \right)dx} \\ = \frac{{{x^{2 + 1}}}}{{2 + 1}} - 2{\rm{x}} + \frac{{{x^{ - 2 + 1}}}}{{ - 2 + 1}} + C = \frac{{{x^3}}}{3} - 2{\rm{x}} - \frac{1}{x} + C\end{array}$.

e)

$\int {\frac{{\left( {x - 3} \right)\left( {x + 1} \right)}}{x}dx}  = \int {\frac{{{x^2} - 2{\rm{x}} - 3}}{x}dx}  = \int {\left( {x - 2 - \frac{3}{x}} \right)dx}  = \frac{{{x^2}}}{2} - 2{\rm{x}} - 3\ln \left| x \right| + C$.

g)

$\begin{array}{l}\int {\left( {3{{\rm{x}}^2} - \frac{4}{x}} \right)\left( {2{\rm{x}} + 5} \right)dx}  = \int {\left( {6{{\rm{x}}^3} + 15{{\rm{x}}^2} - 8 - \frac{{20}}{x}} \right)dx} \\ = \frac{{6{{\rm{x}}^4}}}{4} + \frac{{15{{\rm{x}}^3}}}{3} - 8{\rm{x}} - 20\ln \left| x \right| + C = \frac{3}{2}{{\rm{x}}^4} + 5{{\rm{x}}^3} - 8{\rm{x}} - 20\ln \left| x \right| + C\end{array}$.