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Mathematics
Calculus
Challenging Integrals in Calculus 1-2: Expand Your Problem-Solving Skills!
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[QUOTE="DyslexicHobo, post: 1356850, member: 71687"] I'm just out of Calc 1, so I'm not sure if I even have the knowledge to solve this... but here's where I am now. I can't tell if I complicated it even more, or if I'm closer to getting the solution. [tex]\int (r-1)^{1/5}ln(r) - r^{-1} (r-1)^{1/5} dr[/tex] I used u-substitution (well, r-substitution), where [tex]r = x^5 + 1[/tex]. After the substitution I used integration by parts, and now I'm unsure if that was even the right path. Please let me know! And please tell us how to do [tex]\int_{0}^{\infty}sin(x^2)dx[/tex] [/QUOTE]
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Mathematics
Calculus
Challenging Integrals in Calculus 1-2: Expand Your Problem-Solving Skills!
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