Integrate 3 problems? 1.)(5/ 4x +3) dx 2.) (4x/ sqrt x^2 - 4) dx 3.) (e^cosx) sinx dx
by 
spudd9255
ANSWERS: 2
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The first one is 5/4(ln(x+3/4) .. Second one is I need pen and paper(sorry dont have) third one is -e^cosx hope that helps. Come back to me if you ned the second one.
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You need to put more brackets in when you write in text: 5/4x+3 could be (5/4x) + 3 or 5/(4x + 3) I think you mean 4/(4x+3) In the second one I assume 4x/sqrt(x^2 + 4) To solve: Use change of variable. Change of variable uses one of these rules: Integral f(x) dx = Integral g(u) (dx/du) du Integral f(x) (du/dx) dx = Integral g(u) du where g(u) is the formula f(x) as it would have to be written if you used u instead of x. So in the first one you put u=4x+3 and dx/du = (1/4) So the integral becomes (5/4) Integral (1/u) du which is one of the standard integrals. The second one you use u = sqrt(x^2-4) du/dx = x/u So Integral (4x/sqrt(x^2- 4) dx = Integral 4(x/u) dx = Integral 4(du/dx)dx = Integral 4 du = 4 u + C = 4 sqrt(x^2 - 4) + C Check by differentiating. In the third one you use u = cos x, du /dx = -sin x so Integral (e^cos x)(sin x)dx = Integral e^u (-(du/dx)) dx = - Integral e^u (du/dx) dx = - Integral e^u du = - e^u + C = - e^(cos x) + C
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