Integral dx/x^2+4x+2 - Universityku

Integral dx/x^2+4x+2

Jawaban Terkonfirmasi

$displaystyle intfrac{1}{x^2+4x+2},dx=intfrac{1}{(x+2)^2-2},dx\intfrac{1}{x^2+4x+2},dx=intfrac{1}{(x+2+sqrt2)(x+2-sqrt2)},dx\intfrac{1}{x^2+4x+2},dx=intfrac{1}{2sqrt2(x+2-sqrt2)}-frac{1}{2sqrt2(x+2+sqrt2)},dx\intfrac{1}{x^2+4x+2},dx=frac1{2sqrt2}intfrac{1}{x+2-sqrt2}-frac{1}{x+2+sqrt2},dx\intfrac{1}{x^2+4x+2},dx=frac1{2sqrt2}left(ln|x+2-sqrt2|-ln|x+2+sqrt2|right)+C\boxed{boxed{intfrac{1}{x^2+4x+2},dx=frac{ln|x+2-sqrt2|-ln|x+2+sqrt2|}{2sqrt2}+C}}$