Subiectul III, ex.2.c.

Se consideră funcţia \(f:(0,+\infty)\rightarrow (0,+\infty),\;\;f(x)=x+\frac{9}{x}\).
Determinaţi numărul real \(a\) , ştiind că \(\int_1^{\sqrt{3}}\left(f(x)-\frac{9}{x}\right)arctg\; x\;dx=\frac{5\pi}{12}-\frac{3+\sqrt{3}-a}{2}\).
Soluţie. \begin{align} &\cssId{Step1}{\int_1^{\sqrt{3}}\left(f(x)-\frac{9}{x}\right)arctg\;x\;dx=\int_1^{\sqrt{3}}\left(x+\frac{9}{x}-\frac{9}{x}\right)arctg\; x\;dx=}\\ &\cssId{Step2}{=\int_1^{\sqrt{3}}x\cdot arctg\;x\;dx=\int_1^{\sqrt{3}}arctg\; x\;\cdot\left(\frac{x^2+1}{2}\right)^,dx= }\\ &\cssId{Step3}{=\left(\frac{x^2+1}{2}arctg\; x\right)\mid_1^{\sqrt{3}}-\int_1^{\sqrt{3}}\frac{x^2+1}{2(x^2+1)}dx=}\\ &\cssId{Step4}{=\left(\frac{x^2+1}{2}arctg\; x-\frac{x}{2}\right)\mid_1^{\sqrt{3}}=}\\ &\cssId{Step5}{=\frac{(\sqrt{3})^2+1}{2}arctg\;\sqrt{3}-\frac{\sqrt{3}}{2}-\frac{1^2+1}{2}arctg\;1+\frac{1}{2}=}\\ &\cssId{Step6}{=\frac{4}{2}\cdot\frac{\pi}{3}-\frac{\sqrt{3}}{2}-\frac{\pi}{4}+\frac{1}{2}=\frac{2\pi}{3}-\frac{\pi}{4}-\frac{\sqrt{3}-1}{2}=}\\ &\cssId{Step7}{=\frac{8\pi-3\pi}{12}-\frac{\sqrt{3}-1}{2}=\frac{5\pi}{12}-\frac{\sqrt{3}-1}{2}}\\ &\cssId{Step8}{\Rightarrow \frac{\sqrt{3}-1}{2}=\frac{3+\sqrt{3}-a}{2} }\\ &\cssId{Step9}{\Rightarrow -1=3-a\Rightarrow a=4 }\\ \end{align}