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# Hard integral - maybe

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1. Without using substitution, evaluate the integral
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2. (Original post by Ano123)
Without using substitution, evaluate the integral
.
Are you asking a question or posing a challenge? In any case multiply the denominator and numerator by so the integrand becomes . Then use the relation to evaluate the integral.
3. Spoiler:
Show
Edit: Damn I always get ninjaed by someone more elegant.

4. Ok. A harder one I think.

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5. (Original post by Ano123)
Ok. A harder one I think.

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Substitute ; it becomes easy partial fractions after that.
6. (Original post by Zacken)
Substitute ; it becomes easy partial fractions after that.
I used sub . I made this question up and didn't see the sun you made.
Did you get an answer or just spot the substitution without actually doing it?

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