$\int_{\frac{1}{3}}^{1}\frac{(x-x^{3})^{\frac{1}{3}}}{x^{4}}dx$
$\int_{\frac{1}{3}}^{1}\frac{(x-x^{3})^{\frac{1}{3}}}{x^{4}}dx$
Started By thuanhoang1712, 16-04-2012 - 21:34
#1
Posted 16-04-2012 - 21:34
#2
Posted 17-04-2012 - 13:57
$\int_{\frac{1}{3}}^1\frac{(x-x^3)^{\frac{1}{3}}}{x^4}{\rm d}x=\int_{\frac{1}{3}}^1\frac{(\frac{1}{x^2}-1)^{\frac{1}{3}}}{x^3}{\rm d}x$
Đặt $t=\frac{1}{x^2}-1$ suy ra ${\rm d}t=-\frac{1}{2x^3}{\rm d}x$.
Ta có, $I=\int_{\frac{1}{3}}^1\frac{(\frac{1}{x^2}-1)^{\frac{1}{3}}}{x^3}{\rm d}x=2\int_{0}^8t^{\frac{1}{3}}{\rm d}t=2.\frac{3}{4}t^{\frac{4}{3}}\Bigg|_0^8=24$
Đặt $t=\frac{1}{x^2}-1$ suy ra ${\rm d}t=-\frac{1}{2x^3}{\rm d}x$.
Ta có, $I=\int_{\frac{1}{3}}^1\frac{(\frac{1}{x^2}-1)^{\frac{1}{3}}}{x^3}{\rm d}x=2\int_{0}^8t^{\frac{1}{3}}{\rm d}t=2.\frac{3}{4}t^{\frac{4}{3}}\Bigg|_0^8=24$
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