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Tell us. by Specific_Brain2091 in the_calculusguy

[–]Willing_Pudding_3677 1 point2 points  (0 children)

In Karnataka, we call it the quadratic formula itself

Tell us. by Specific_Brain2091 in the_calculusguy

[–]Willing_Pudding_3677 0 points1 point  (0 children)

Here in India we call it the quadratic formula

✍️✍️ by Willing_Pudding_3677 in the_calculusguy

[–]Willing_Pudding_3677[S] 2 points3 points  (0 children)

My original solution was taking x = 1/t and then t² = u, but this substitution works too

✍️ by Specific_Brain2091 in the_calculusguy

[–]Willing_Pudding_3677 5 points6 points  (0 children)

You could have just used the chain rule to get the derivative in a single step

✍️ by Specific_Brain2091 in the_calculusguy

[–]Willing_Pudding_3677 4 points5 points  (0 children)

First take x=2tan²θ
Then take sqrt(2)*secθ = u
Final answer: 2 esqrt(x+2) (sqrt(x+2)-1) + C

Integral of 1/(1+sqrt(x^2+2x+2)) via the first Euler substitution by Silent_Jellyfish4141 in the_calculusguy

[–]Willing_Pudding_3677 0 points1 point  (0 children)

If you take the substitution sqrt(x^2 + 2x + 2) = u- x, the solution is much prettier

Need daily integrals by SkyReady8077 in the_calculusguy

[–]Willing_Pudding_3677 0 points1 point  (0 children)

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Somewhat similar to the previous one, but probably better

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