Giải Luyện tập 5 trang 10 SGK Toán 12

Lời giải

a) $\displaystyle\int \dfrac{1}{x^4}\,dx = \int x^{-4}\,dx = \dfrac{x^{-4+1}}{-4+1} + C = \dfrac{x^{-3}}{-3} + C = -\dfrac{1}{3x^3} + C.$

Kiểm tra: $\left(-\dfrac{1}{3}x^{-3}\right)' = -\dfrac{1}{3}\cdot(-3)x^{-4} = x^{-4} = \dfrac{1}{x^4}.$ ✓

b) Với $x>0$: $x\sqrt{x} = x\cdot x^{\frac12} = x^{\frac32}.$

$\displaystyle\int x\sqrt{x}\,dx = \int x^{\frac32}\,dx = \dfrac{x^{\frac32+1}}{\frac32+1} + C = \dfrac{x^{\frac52}}{\frac52} + C = \dfrac{2}{5}x^{\frac52} + C = \dfrac{2}{5}x^2\sqrt{x} + C.$

Kiểm tra: $\left(\dfrac{2}{5}x^{\frac52}\right)' = \dfrac{2}{5}\cdot\dfrac{5}{2}x^{\frac32} = x^{\frac32} = x\sqrt{x}.$ ✓

c) Với $x>0$: $\sqrt[3]{x} = x^{\frac13}.$

$\displaystyle\int\left(\dfrac{3}{x} - 5\sqrt[3]{x}\right)dx = 3\int\dfrac{1}{x}\,dx - 5\int x^{\frac13}\,dx.$

$\displaystyle = 3\ln x - 5\cdot\dfrac{x^{\frac13+1}}{\frac13+1} + C = 3\ln x - 5\cdot\dfrac{x^{\frac43}}{\frac43} + C = 3\ln x - \dfrac{15}{4}x^{\frac43} + C = 3\ln x - \dfrac{15}{4}x\sqrt[3]{x} + C.$

Kiểm tra: $\left(3\ln x - \dfrac{15}{4}x^{\frac43}\right)' = \dfrac{3}{x} - \dfrac{15}{4}\cdot\dfrac{4}{3}x^{\frac13} = \dfrac{3}{x} - 5x^{\frac13} = \dfrac{3}{x} - 5\sqrt[3]{x}.$ ✓

Đáp số