Question

integration of x^8/(1-x^3)^1/3​

Answer 1

Answer:

$\displaystyle\tfrac1{40}(1-x^3)^{2/3}(9+6x^3+5x^6) + C$

Step-by-step explanation:

Make the substitution

• u³ = 1 - x³

Then

• u⁶ = (1 - x³)² = 1 - 2x³ + x⁶
• x³ = 1 - u³
• x⁶ = (1 - u³)² = 1 - 2u³ + u⁶

• 3u² du = -3x² dx  ⇒  u² du = -x² dx

So...

$\displaystyle\int\frac{x^8\,dx}{(1-x^3)^{1/3}}\\\\=-\int\frac{x^6\times(-x^2\,dx)}{(1-x^3)^{1/3}}\\\\=-\int\frac{(1-2u^3+u^6)\times u^2\,du}{u}\\\\=\int(-1+2u^3-u^6)\times u\,du\\\\=\int(-u+2u^4-u^7)\,du\\\\=-\tfrac12u^2+\tfrac25u^5-\tfrac18u^8 + C\\\\=-\tfrac1{40}u^2(20-16u^3+5u^6)+C$

$\displaystyle=-\tfrac1{40}(1-x^3)^{2/3}\bigl(20-16(1-x^3)+5(1-2x^3+x^6)\bigr)+C\\\\=\tfrac1{40}(1-x^3)^{2/3}(20-16+16x^3+5-10x^3+5x^6)+C\\\\=\tfrac1{40}(1-x^3)^{2/3}(9+6x^3+5x^6) + C$

Hope that helps.