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## Re: integral calculus logarithmic functions

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

## Re: integral calculus logarithmic functions

In reply to Re: integral calculus logarithmic functions by Jhun Vert

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

## Re: integral calculus logarithmic functions

In reply to Re: integral calculus logarithmic functions by Jeanvill Palad…

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

## Re: integral calculus logarithmic functions

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 x

_{2}: 0 = A + B, B = -1Equate 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.