when using integration tables, how important is the constant in the original integrad?
I have the problem int sqrt(7-4x-4x^2) dx.
i've found properties for ax^2+-b. but nothing that has ax^2 bx and c.
i have to imagine that the 7 just doesn't disappear.
and i tried factoring it to change how it looks to int sqrt(4x)sqrt(7-(1-x)) dx but that dosen't seem to help much either.
I have the problem int sqrt(7-4x-4x^2) dx.
i've found properties for ax^2+-b. but nothing that has ax^2 bx and c.
i have to imagine that the 7 just doesn't disappear.
and i tried factoring it to change how it looks to int sqrt(4x)sqrt(7-(1-x)) dx but that dosen't seem to help much either.
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∫√(7 - 4x - 4x^2) dx =
∫√(8 -1 - 4x - 4x^2) dx =
∫√(8 - (2x + 1) ^2) dx =
use u = 2x+1 and du = 2dx and 8 = a^2 you get
(1/2) ∫ √(a^2 - u^2) du= (1/4)[ u√(a^2 - u^2) + (a^2)*Arcsin(u/√(a^2 - u^2))] + C
Now back to x
(1/4)[ (2x + 1)√(8 - (2x + 1)^2) + 8Arcsin(u/√(8 - (2x + 1)^2))] + C
∫√(8 -1 - 4x - 4x^2) dx =
∫√(8 - (2x + 1) ^2) dx =
use u = 2x+1 and du = 2dx and 8 = a^2 you get
(1/2) ∫ √(a^2 - u^2) du= (1/4)[ u√(a^2 - u^2) + (a^2)*Arcsin(u/√(a^2 - u^2))] + C
Now back to x
(1/4)[ (2x + 1)√(8 - (2x + 1)^2) + 8Arcsin(u/√(8 - (2x + 1)^2))] + C
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The 7 isn't a constant. It's inside the square root and thus an integral part of the function. Also, to fit the form you've given you'd have to dismiss the -4x, not the 7.
#37 on this list is the one you want:
http://integral-table.com/integral-table…
#37 on this list is the one you want:
http://integral-table.com/integral-table…