# integral calculus logarithmic functions

5 posts / 0 new
annnarose
integral calculus logarithmic functions

Pano po naging sagot sa problem na integral of dx all over x(1+x^2) e 1/2 ln x^2 over 1+x^2 +c

Jhun Vert

hindi po malinaw kung ano talaga ang equation. You can use symbols of grouping (brackets) para mas madali maintindihan.

patulong nga po. 3x^4+4x^3+16x^2+20x+9 over (x+20)(x^2+3)^2 pasagot nga po yan hirap po kase

Jhun Vert

Ano ang tanong dito? Mahirap manghula kuna ano ang ibig mong sabihin.

Jhun Vert

Binasa ko uli ang post mo at naintindihan ko na. You ask why
$\displaystyle \int \dfrac{dx}{x(1 + x^2)} = \dfrac{1}{2} \, \ln \left( \dfrac{x^2}{1 + x^2} \right) + C$

am I right? If so, here is the detail:

$\dfrac{1}{x(1 + x^2)} = \dfrac{A}{x} + \dfrac{Bx + C}{1 + x^2}$

$1 = A(1 + x^2) + Bx^2 + Cx$

When x = 0, A = 1
Equate x2: 0 = A + B, B = -1
Equate x: C = 0

Thus,
$\displaystyle \int \dfrac{dx}{x(1 + x^2)} = \int \left( \dfrac{1}{x} - \dfrac{x}{1 + x^2} \right) \, dx$

$\displaystyle \int \dfrac{dx}{x(1 + x^2)} = \int \dfrac{dd}{x} - \dfrac{1}{2} \int \dfrac{2x \, dx}{1 + x^2}$

$\displaystyle \int \dfrac{dx}{x(1 + x^2)} = \ln x - \dfrac{1}{2} \ln (1 + x^2) + C$

$\displaystyle \int \dfrac{dx}{x(1 + x^2)} = \dfrac{1}{2} \left[ 2\ln x - \ln (1 + x^2) \right] + C$

$\displaystyle \int \dfrac{dx}{x(1 + x^2)} = \dfrac{1}{2} \left[ \ln x^2 - \ln (1 + x^2) \right] + C$

$\displaystyle \int \dfrac{dx}{x(1 + x^2)} = \dfrac{1}{2} \, \ln \left( \dfrac{x^2}{1 + x^2} \right) + C$

as we expect it to be.

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