1) Odrediti integral
\[ \int \frac{\arctan(2x)}{(2x-1)^{2}}\,dx. \]Rešenje
Parcijalna integracija sa \(u=\arctan(2x)\), \(dv=(2x-1)^{-2}\,dx\) daje \[ \int \frac{\arctan(2x)}{(2x-1)^{2}}\,dx = -\frac{\arctan(2x)}{2x-1} + \int \frac{2}{(1+4x^2)(2x-1)}\,dx. \]Rastavljanje razlomka: \[ \frac{2}{(1+4x^2)(2x-1)} = \frac{A}{2x-1}+\frac{Bx+C}{\,1+4x^2\,}. \] Množenjem imeniocem \((1+4x^2)(2x-1)\): \[ 2=A(1+4x^2)+(Bx+C)(2x-1) =(4A+2B)x^2+(2C-B)x+(A-C). \] Upoređivanjem koeficijenata: \(A=1,\ B=-2,\ C=-1\). Zato \[ \frac{2}{(1+4x^2)(2x-1)} = \frac{1}{2x-1}-\frac{2x+1}{1+4x^2}. \]
Integrali: \[ \int \frac{1}{2x-1}\,dx=\left(\frac{1}{2}\right)\ln|2x-1|,\quad \int \frac{2x}{1+4x^2}\,dx=\left(\frac{1}{4}\right)\ln(1+4x^2),\quad \int \frac{1}{1+4x^2}\,dx=\left(\frac{1}{2}\right)\arctan(2x). \] Dakle \[ \int \frac{2}{(1+4x^2)(2x-1)}\,dx = \left(\frac{1}{2}\right)\ln|2x-1|-\left(\frac{1}{4}\right)\ln(1+4x^2)-\left(\frac{1}{2}\right)\arctan(2x). \]
Rezultat: \( -\frac{\arctan(2x)}{2x-1} + \left(\frac{1}{2}\right)\ln|2x-1| – \left(\frac{1}{4}\right)\ln(1+4x^2) – \left(\frac{1}{2}\right)\arctan(2x) + C \).2) Izračunati:
\[ \int_{1}^{e^{2}} \frac{\sqrt{\,8\ln x – (\ln x)^2\,}}{x}\,dx. \]Rešenje
Neka je \(u=\ln x\Rightarrow du=dx/x\), granice: \(x\in[1,e^2]\Rightarrow u\in[0,2]\). \[ \int_{1}^{e^{2}} \frac{\sqrt{8\ln x-(\ln x)^2}}{x}\,dx =\int_{0}^{2}\!\sqrt{8u-u^2}\,du =\int_{0}^{2}\!\sqrt{16-(u-4)^2}\,du. \]
Smena: \(u-4=4\sin t \Rightarrow du=4\cos t\,dt\). Granice: \(u=0\Rightarrow t=-\pi/2\), \(u=2\Rightarrow t=-\pi/6\). Tada \[ \sqrt{16-(u-4)^2}=4\cos t \quad (\cos t\ge0 \text{ na } [-\pi/2,-\pi/6]). \] Zato \[ I=\int_{-\pi/2}^{-\pi/6} 4\cos t\cdot 4\cos t\,dt =16\int_{-\pi/2}^{-\pi/6}\!\cos^2 t\,dt. \]
\(\cos^2 t=\frac{1+\cos 2t}{2}\) ⇒ \[ 16\int\cos^2 t\,dt =16\int \frac{1+\cos 2t}{2}\,dt =8t+4\sin 2t. \]
Rezultat: \( (8\pi)/3 – 2\sqrt{3} \).Zamenom granica:
\[ I=\left.\,8t+4\sin(2t)\,\right|_{t=-\pi/2}^{\,t=-\pi/6} =\frac{8\pi}{3}-2\sqrt{3}. \]© TMF/Matematika 2