Question

как?! помогите, прошу

Answer 1

$intlimits^5_1, Big(dfrac{3}{2sqrt{3x+1}}cdot lnx+dfrac{sqrt{3x+1}}{x}Big), dx=intlimits^5_1, dfrac{3cdot lnx}{2sqrt{3x+1}}, dx+intlimits^5_1, dfrac{sqrt{3x+1}}{x}, dx=Q$

$star ; int dfrac{3cdot lnx}{2sqrt{3x+1}}, dx=Big [; u=3, lnx; ,; du=frac{3, dx}{x}; ,; dv=frac{dx}{2sqrt{3x+1}}; ,; v=frac{sqrt{3x+1}}{3}; Big]=\\=uv-int v, du=sqrt{3x+1}cdot lnx-int dfrac{sqrt{3x+1}}{x}, dx; ;$

$Q=sqrt{3x+1}cdot lnxBig |_1^5-intlimits^5_1dfrac{sqrt{3x+1}}{x}, dx+intlimits^5_1dfrac{sqrt{3x+1}}{x}, dx=sqrt{3x+1}cdot lnxBig |_1^5=\\\=sqrt{16}cdot ln5-sqrt4cdot ln1=4, ln5-0=4, ln5$

$P.S.; int dfrac{sqrt{3x+1}}{x}, dx=Big[; t=sqrt{3x+1}; ,; t^2=3x+1; ,; x=frac{1}{3}(t^2-1); ,\\dx=frac{2t}{3}, dt; Big]=int dfrac{3t}{t^2-1} cdot dfrac{2t}{3}, dt=2int dfrac{t^2, dt}{t^2-1}=2int Big(1+dfrac{1}{t^2-1}Big), dt=\\=2Big (t+dfrac{1}{2}, lnBig|dfrac{t-1}{t+1}Big|Big)+C=2sqrt{3x+1}+lnBig |dfrac{sqrt{3x+1}-1}{sqrt{3x+1}+1}Big|+C$