Question

Seja F a primitiva da funçao f(x)= x+4/ x^2-12x+35 tal que F(6) = 8. Calcule o valor de F(0). Imagem Anexada

Answer 1

$\displaystyle \sf f(x) =\frac{x+4}{x^2-12x+35} \\\\\\ \text{Primitiva da f(x)} : \\\\ F(x) = \int \frac{x+4}{x^2-12x+35} \ dx \\\\\\\ F(x) = \int \frac{x+4}{x^2-12x+36-1}\ dx \\\\\\ F(x) =\int \frac{x+4}{(x-6)^2-1}\ dx \\\\\\ \text{Fa{\c c}amos : } \\\\ u = x-6 \to du = dx \\\\ u+10 = x+4 \\\\ \text{Da{\'i}}: \\\\ F(x) =\int \frac{u+10}{u^2-1}\ du \\\\\\ F(x) = \int \frac{u}{u^2-1}\ du+\int \frac{10}{u^2-1} \ du$

$\displaystyle \sf F(x) = \frac{1}{2}\int \frac{2u}{u^2-1}\ du + \int \frac{10}{u^2-1} \ du \\\\\\ F(x) = \frac{1}{2}\cdot \ln\left |u^2-1\right | + \int\frac{10}{(u+1)(u-1)} \ du \\\\\\ \text{Fra{\c c}{\~a}o parcial} : \\\\ \frac{10}{(u+1)(u-1)} = \frac{A}{u+1}+\frac{B}{u-1} \\\\\\ \frac{10}{(u+1)(u-1)} =\frac{A(u-1)+B(u+1)}{(u+1)(u-1)} \\\\\\ \frac{10}{(u+1)(u-1)} =\frac{Au-A+Bu+B}{(u+1)(u-1)}\\\\\\ \frac{10}{(u+1)(u-1)} =\frac{(A+B)u+B-A}{(u+1)(u-1)}$

$\left\{ \begin{array}{I} \displaystyle \sf A+B = 0 \to A=-B \\\\ \displaystyle \sf B-A = 10\to 2B = 10\\\\ \displaystyle \sf B = 5 \to A = -5 \end{array}$

$\displaystyle \sf Da{\'i}} : \\\\ \ F(x) = \frac{1}{2}\cdot \ln \left|u^2-1 \right| + \int \frac{-5}{u+1}\ du + \int \frac{5}{u-1} \ du \\\\\\ F(x ) =\frac{1}{2}\cdot \ln \left|u^2-1 \right| -5\cdot \ln \left| u+1\right|+5\cdot \ln\left|u-1\right| + C \\\\\\\ F(x) = \frac{1}{2}\cdot \ln \left|(x-6)^2-1 \right|-5\cdot \ln \left| x-6+1\right|+5\cdot \ln \left| x-6-1\right| + C\\\\\\ F(x) = \frac{1}{2}\cdot \ln \left|(x-6)^2-1\right|-5\ln\left|x-5\right| + 5\ln\left|x-7\right| + C$

$\displaystyle \sf \text{Fa{\c c}amos} :\\\\ F(6) = 8 \\\\ 8 = \frac{1}{2} \cdot \ln |-1| -5\ln |1|+5\ln|-1| + C \\\\\ C = 8 \\\\ Da{\'i}} : \\\\\ F(x) = \frac{1}{2}\cdot \ln \left|(x-6)^2-1\right|-5\ln\left|x-5\right| + 5\ln\left|x-7\right|+8 \\\\\\ F(0) = \frac{1}{2}\ln |35|-5\ln|-5|+5\ln|-7|+8\\\\\\ F(0)=\frac{1}{2}\cdot 3,5553-5\cdot 1,6094+5\cdot 1,9459+8\\\\ F(0) = 1,777-8,047+9,729+8\\\\ \huge\boxed{\sf F(0) = 11,45 }\checkmark$