Oblicz całki: a) ∫12x+8/4x²+1 dx b) ∫cos²xsinxdx

a) [latex] intlimits { dfrac{12x+8}{4x^{2}+1} } , dx = intlimits { dfrac{ frac{3}{2}(8x + frac{16}{3}) }{4x^{2}+1} } , dx = dfrac{3}{2} intlimits { dfrac{8x + frac{16}{3}}{4x^{2}+1} } , dx = \ \ = dfrac{3}{2} left ( intlimits { dfrac{8x}{4x^{2}+1} } , dx + intlimits { dfrac{ frac{16}{3} }{4x^{2}+1} } , dx ight ) = \ \ = dfrac{3}{2} left ( intlimits { dfrac{8x}{4x^{2}+1} } , dx + dfrac{16}{3} intlimits { dfrac{1}{4x^{2}+1}} } , dx ight )[/latex] Pierwsza z całek. Widzimy, że licznik jest pochodną mianownika, zatem : [latex]intlimits { dfrac{8x}{4x^{2}+1} } , dx = ln |4x^{2}+1| + C =oxed{ ln left (4x^{2}+1 ight ) + C}[/latex] Teraz druga całka. Zastosujemy podstawienie : [latex]intlimits { dfrac{1}{4x^{2}+1}} } , dx = left|egin{array}{ccc} t = 2x \ dt = 2dx \ dx = frac{dt}{2} end{array} ight| = intlimits { dfrac{ frac{dt}{2} }{t^{2}+1} } , = dfrac{1}{2} intlimits { dfrac{dt}{t^{2}+1} } , = \ \ = dfrac{1}{2} arctan t + C = oxed{ dfrac{1}{2} arctan (2x) + C } [/latex] Ostatecznie wynik to : [latex] intlimits { dfrac{12x+8}{4x^{2}+1} } , dx = dfrac{3}{2} left [ ln left (4x^{2}+1 ight ) + dfrac{16}{3} cdot dfrac{1}{2} arctan (2x) ight ] = \ \ = oxed{oxed{ dfrac{3}{2} ln left (4x^{2} + 1 ight ) + 4 arctan(2x) }}[/latex] ========================== Stosujemy podstawienie. [latex] intlimits {cos^{2}x sin x} , dx = left|egin{array}{ccc} t = cos x \ dt = - sin x dx \ sin dx = - dt end{array} ight| = intlimits {t^{2}} , (-dt) = \ \ = - intlimits {t^{2}} , dt = - dfrac{1}{3}t^{3} + C = oxed{ - dfrac{1}{3}cos^{3} x + C } [/latex]