Integral calculus
Hello, this is the last question from my homework that I get stuck at:
∫ (5^3/x - (x^8 + 9)^1/3 ) / 3x^2
will u, for sub. be = x^8 + 9?
Then du is = 8x^7 dx
x^7 dx = 1/8 du
from there I'm stuck.
Thank you
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Answers
The substitution u = x^8+9 does indeed work, and you have correctly evaluated the differential du as 8x^7dx. To continue the solution, you need to use the chain rule to rewrite the original integrand in terms of u. Note that: du = 8x^7dx du/dx = 8x^7 dx/du = 1/8x^(-7) Therefore, you can rewrite the integrand as: (5^3/x - (x^8 + 9)^1/3)/3x^2 = (5^3/x - u^1/3) * (1/8x^(-7))/(3x^2) = (5^3/8x^8 - u^1/3*1/8u^(-7))/(3*1/8u^(-7)) = (5^3u^7 - u^1/3)/(24u^7) Now that everything is in terms of u we can proceed with the integration. ∫ (5^3u^7 - u^1/3) / (24u^7) du = 1/24 (5^3u^
