Obliczyć całki nieoznaczone. (2 przykłady)

[zaqws]

a) \(\int_{}^{} \frac{ \arcsin \sqrt{x} }{x^2} dx\)
b) \(\int_{}^{} \frac{dx}{(3x^2 + 1)^2}\)

[panb]

a) \(\int \frac{\arcsin\sqrt x}{x^2}dx= \begin{vmatrix}t=\sqrt x& x^2=t^4\\dt= \frac{dx}{2\sqrt x}& dx=2\sqrt x dt=2tdt \end{vmatrix}=\int \frac{\arcsin t \cdot 2t}{t^4} dt=2\int \frac{\arcsin t}{t^3}dt\\
2\int \frac{\arcsin t}{t^3}dt= \begin{vmatrix}u=\arcsin t & du =\frac{dt}{\sqrt{1-t^2}}\\dv=\frac{dt}{t^3}&v= - \frac{1}{2t^2} \end{vmatrix}=2 \left[- \frac{1}{2} \frac{\arcsin t}{t^2}+ \frac{1}{2} \int \frac{dt}{t^2\sqrt{1-t^2}} \right]
=-\frac{\arcsin t}{t^2}+\int\frac{dt}{t^2\sqrt{1-t^2}}\\
\int\frac{dt}{t^2\sqrt{1-t^2}}= \begin{vmatrix} t=\sin u & dt=\cos u du\\ \sqrt{1-t^2}=\cos u\end{vmatrix}=\int \frac{\cos u }{\sin^2u \cdot \cos u}du=\int \frac{du}{\sin^2u} =-\ctg u= \\=-\frac{\cos u}{\sin u}= -\frac{\sqrt{1-\sin^2 u}}{\sin u}=- \frac{\sqrt{1-t^2}}{t}\)
Ostatecznie

• \(\int \frac{\arcsin\sqrt x}{x^2}dx =- \frac{\arcsin \sqrt x}{x}- \frac{\sqrt{1-x}}{\sqrt x}+C =-\frac{\arcsin \sqrt x+\sqrt{x-x^2}}{x}+C\)

[panb]

b) \(\int \frac{dx}{(3x^2+1)^2}= \begin{vmatrix} x= \frac{\tg t}{\sqrt3} \So dx= \frac{dt}{\sqrt3 \cos^2t} \wedge t=\arctg(x\sqrt3)\\3x^2+1=3 \cdot \frac{\tg^2t}{3}+1=\frac{1}{\cos^2t} \end{vmatrix}= \frac{1}{\sqrt3}\int \frac{\cos^4t}{\cos^2t}dt= \frac{1}{\sqrt3}\int \cos^2t dt= \frac{1}{\sqrt3}\int \left(\cos2t+\frac{1}{2} \right)dt=\\
= \frac{\sin2t}{4\sqrt3}+\frac{t}{2\sqrt3}= \frac{2\sin t\cos t}{4\sqrt3}+\frac{\arctg(x\sqrt3)}{2\sqrt3} = \frac{\tg t \cdot \cos^2t}{2\sqrt3}+\frac{\arctg(x\sqrt3)}{2\sqrt3}=\frac{x\sqrt3}{2\sqrt3 \cdot (3x^2+1)}+\frac{\arctg(x\sqrt3)}{2\sqrt3}\)

Ostatecznie

• \(\int \frac{dx}{(3x^2+1)^2}= \frac{3x+\sqrt3(3x^2+1)\arctg (x\sqrt3)}{6(3x^2+1)}+C\)

Ufff...