What is the antiderivite of x/(sq. rt.(16+x^2))?

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2007-04-30 14:08:52 · 3 answers · asked by behindthearc32 1 in Science & Mathematics ➔ Mathematics

∫x / √(16 + x^2) dx: let u = 16 + x^2, du = 2x dx.
= ∫(1/2) du / √u
= ∫(1/2)u^(-1/2) du
= (1/2) u^(1/2) / (1/2) + c
= √u + c
= √(16 + x^2) + c.

2007-04-30 14:20:15 · answer #1 · answered by Scarlet Manuka 7 · 0⤊ 0⤋

its 2(16+x^2)^1/2

2007-04-30 21:21:20 · answer #2 · answered by jinfeng l 2 · 0⤊ 0⤋

If you put 16+x^2=z
2xdx=d z so you have
INT1/2*1/sqrt z dz = sqrt z +C = sqrt(16+x^2)+c

2007-04-30 21:21:07 · answer #3 · answered by santmann2002 7 · 0⤊ 0⤋