1. \( \int x \cdot \mathrm{arcctg}^{2} x d x \)

$$ \begin{array}{ll} u=\mathrm{arcctg}^{2} x & d v=x d x \\ d u=2 \mathrm{arcctg} x \cdot \frac{-1}{1+x^{2}}d x & v=\frac{x^{2}}{2} \end{array} $$ \begin{align} I&=\mathrm{arcctg}^{2} x \cdot \frac{x^{2}}{2}+\int \frac{x^{2}}{2} \cdot 2 \mathrm{arcctg} x \cdot \frac{1}{1+x^{2}} dx \\ I&=\frac{x^{2}}{2} \mathrm{arcctg}^{2} x +\underbrace{\int \frac{x^{2} \mathrm{arcctg} x}{1+x^{2}} d x}_{I_{1}} \\ \end{align}

Primenimo sada parcijalnu integraciju na integral \(\quad I_{1}=\int \frac{x^{2} \mathrm{arcctg} x}{1+x^{2}} d x\)

$$ \begin{array}{ll} u=\mathrm{arcctg} x & d v=\frac{x^{2}}{1+x^{2}} d x \\ d u=\frac{-1}{1+x^{2}} & v=x-\mathrm{arctg} x \end{array} $$ \begin{align} I_{1}&=\mathrm{arcctg} x \cdot(x-\mathrm{arctg} x)-\int \frac{\mathrm{arctg} x-x}{1+x^{2}} d x \\ I_{1}&=\mathrm{arcctg} x \cdot(x-\mathrm{arctg} x)-\left(\frac{\mathrm{arctg}^{2} x}{2}-\frac{1}{2} \ln \left(1+x^{2}\right)\right)+C \\ \end{align} \[I=\frac{x^{2}}{2}\mathrm{arcctg}{ }^{2} x + \mathrm{arcctg} x \cdot(x-\mathrm{arctg} x)-\frac{1}{2}\mathrm{arctg}^{2} x+\frac{1}{2}\ln \left(1+x^{2}\right)+C\]

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2. Površina koja rotira je ograničena sa krivom i pravom: $$ \begin{aligned} & y=x^{3} \\ & x-y=0 \Rightarrow y=x. \end{aligned} $$ Kako je $$ \begin{aligned} & x^{3}=x \\ & x^{3}-x=0 \\ & x(x-1)(x+1)=0 \\ & x=1 \quad x=0 \quad x=-1 \end{aligned} $$ interval na \(x\)-osi je \([0,1]\).

Obrtanjem oko \(x\)-ose formira se površ \(P_1\) spolja, koju formira prava \(y=x\) (na animaciji braon boje) i površ \(P_2\) unutra, koju formira kriva \(y=x^3\) (na animaciji crne boje).

\(P_{1}=2 \pi \int_{0}^{1} x \cdot \sqrt{1+1^{2}} d x=2 \sqrt{2} \pi \int_{0}^{1} x d x =\left.2 \sqrt{2} \pi \cdot \frac{x^{2}}{2}\right|_{0} ^{1}=\pi \sqrt{2}\)

\[P_{1}=\pi \sqrt{2}\]

\(P_{2}=2 \pi \int_{0}^{1} x^{3} \cdot \sqrt{1+\left(3 x^{2}\right)^{2}} d x,\)

\(\quad\text { smena: } 1+9 x^{4}=t,\quad\left. 36 x^{3} d x=d t,\quad x^{3}d x=d t / 36 \quad\right|_{x=0\; \rightarrow \; t=1}^{x=1\;\rightarrow \; t=10} \)

\(P_{2}=2 \pi \int_{1}^{10} \sqrt{t} \frac{d t}{36}=\left.\frac{\pi}{18} \cdot \frac{t^{3 / 2}}{3 / 2}\right|_{1}^{10} =\frac{\pi}{27}(10 \sqrt{10}-1)\)

\[P_{2}=\frac{\pi}{27}(10 \sqrt{10}-1)\]

$$ \begin{aligned} & P=P_{1}+P_{2} \\ & P=\pi \sqrt{2}+\frac{\pi}{27}(10 \sqrt{10}-1) \end{aligned} $$

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3. Tangenta u tački \(A(6,3)\) krive \(\;y=-x^2+10x-21\;\) ima jednačinu \(\;y-3=k(x-6)\;\) gde je \(\;k=y'(6)=-2\). Jednačina tangente je \(\;y=-2 x+15\).

Prava \(\;10 x+y=15\;\) se može zapisati i kao \(\;y=-10 x+15\).

Najpre računamo površine iznad \(x\)-ose. \begin{align} P_{1}&=\int_{0}^{\frac{3}{2}}(-2 x+15) d x-\int_{0}^{\frac{3}{2}}(-10 x+15) d x=\int_{0}^{\frac{3}{2}} 8 x d x=\left.8 \cdot \frac{x^{2}}{2}\right|_{0} ^{3 / 2}=4 \cdot \frac{9}{4}=9\\ P_{2}&=\int_{\frac{3}{2}}^{3}(-2 x+15) d x=\left.\left(-x^2+15x\right)\right|_{\frac{3}{2}}^{3}=-\left(3^{2}-\left(\frac{3}{2}\right)^{2}\right)+15(3-\frac{3}{2})=-\frac{27}{4}+\frac{45}{2}=\frac{63}{4}\\ P_{3}&=\int_{3}^{6}(-2 x+15) d x-\int_{3}^{6}\left(-x^{2}+10 x-21\right) d x=\int_{3}^{6}\left(x^{2}-12 x+36\right) d x\\ &=\left.\left(\frac{x^{3}}{3}-12 \frac{x^{2}}{2}+36 x\right)\right|_3^6=\frac{6^{3}}{3}-\frac{3^{3}}{3}-6 \left(6^{2}-3^{2}\right)+36(6-3)= 63-162+108=9\\ \end{align} Zatim površine ispod \(x\)-ose. \begin{align} P_{4}&=-\int_{\frac{3}{2}}^{2}(-10 x+15) d x=\left.\left(10\frac{x^2}{2}-15x\right)\right|_{\frac{3}{2}}^2=5\left(2^2-\left(\frac{3}{2}\right)^2\right)-15(2-\frac{3}{2})=5\cdot\frac{7}{4}-\frac{15}{2}=\frac{5}{4} \\ P_{5}&=-\int_{2}^{3}\left(-x^{2}+10 x-21\right) d x=\left.\left(\frac{x^{3}}{3}-10 \frac{x^{2}}{2}+21 x\right)\right|_{2}^{3}=\frac{19}{3}-25+21=\frac{7}{3} \\ \end{align} Na kraju ukupna površina je $$ P=P_1+P_2+P_3+P_4+P_5, $$ $$ P=9+\frac{63}{4}+9+\frac{5}{4}+\frac{7}{3}, $$ $$ P=\frac{112}{3}. $$