Đáp án:

$I=\dfrac{3}{2} \ln (x^2-10x+29)+4\arctan \left(\dfrac{x-5}{2}\right)+C.$

Giải thích các bước giải:

$I=\displaystyle \int \dfrac{3x-7}{x^2-10x+29} \, dx\\ =\dfrac{3}{2}\displaystyle \int \dfrac{2x-\dfrac{21}{2}}{x^2-10x+29} \, dx\\ =\dfrac{3}{2}\displaystyle \int \dfrac{2x-10-\dfrac{1}{2}}{x^2-10x+29} \, dx\\ =\dfrac{3}{2}\displaystyle \int \dfrac{2x-10}{x^2-10x+29} \, dx+8\displaystyle \int \dfrac{1}{x^2-10x+29} \, dx\\ =\dfrac{3}{2}\displaystyle \int \dfrac{ dx(x^2-10x+29)}{x^2-10x+29} \, dx+8I_1\\ =\dfrac{3}{2} \ln (x^2-10x+29)+8I_1+C\\ I_1=\displaystyle \int \dfrac{1}{x^2-10x+29} \, dx\\ =\displaystyle \int \dfrac{1}{(x-5)^2+4} \, dx\\ x-5=2\tan t, dx=2(\tan^2 t+1) \, dt\\ I_1=\displaystyle \int \dfrac{2(\tan^2 t+1)}{4\tan^2t+4} \, dt\\ =\dfrac{1}{2}\displaystyle \int \, dt\\ =\dfrac{1}{2}t\\ =\dfrac{1}{2} \arctan \left(\dfrac{x-5}{2}\right)+C\\ I=\dfrac{3}{2} \ln (x^2-10x+29)+4\arctan \left(\dfrac{x-5}{2}\right)+C.$