The integral of x^3 / (sqrt. (x^2 + 1)) dx.
2006-12-03 13:16:27 · 4 answers · asked by Phosphorus 1 in Science & Mathematics ➔ Mathematics
Let u = x^2 + 1.
Then du/dx = 2x, so dx = du/2x.
So we get the integral of x^3/sqrt(u) * du/2x, which simplifies down to:
0.5x^2/sqrt(u) du.
We can now substitute in for the x^2 as u-1:
0.5(u-1)/sqrt(u) du
and expand and divide:
0.5u^0.5 - 0.5u^-0.5 du.
Integrate:
1/3 u^1.5 - u^0.5 + c
Substitute back in:
1/3 (x^2 + 1)^1.5 - (x^1 + 2)^0.5 + c.
2006-12-03 13:20:53 · answer #1 · answered by stephen m 4 · 0⤊ 0⤋
62
2006-12-03 13:19:24 · answer #2 · answered by Joe Prosnick 5 · 0⤊ 1⤋
it is really x^3/x+1(dx) thats as far as i think i know lol
2006-12-03 13:20:59 · answer #3 · answered by undergroundburn 2 · 0⤊ 1⤋
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2016-10-13 23:02:23 · answer #4 · answered by ? 4 · 0⤊ 0⤋