Całka

[Ewwaa]

Oblicz całkę:

1) \(\int \frac{\sqrt{x}dx}{2+x}\)

2)\(\int\frac{1}{(5+x)\sqrt{x}}dx\)

[sebnorth]

1)

\(\int \frac{\sqrt{x}}{2+x} dx = \int \frac{2\sqrt{x}}{2+x} \cdot \frac{1}{2\sqrt{x}} dx = \ast\)

\(\sqrt{x} = t, \frac{1}{2\sqrt{x}} dx = dt, x = t^2\)

\(\ast = \int \frac{2t^2}{2+t^2} dt = \int \frac{2t^2 + 4 - 4}{t^2 + 2} dt = 2\int dt - 4 \int \frac{dt}{t^2 + 2}\)

\(= 2t - 4 \cdot \frac{1}{2} \int \frac{dt}{( \frac{t}{\sqrt{2}} )^2 + 1} = \ast\)

\(\frac{t}{\sqrt{2}} = u, \frac{1}{\sqrt{2}} dt = du\)

\(\ast = 2t - 2 \int \frac{\sqrt{2}}{u^2 + 1} du = 2t - 2\sqrt{2} \cdot \arctg u + C =\)

\(= 2 \sqrt{x} - 2\sqrt{2} \cdot \arctg \frac{ \sqrt{x} }{ \sqrt{2} } + C\)

[denatlu]

2)

\(\int \frac{dx}{\sqrt{x} \cdot (x+5)} = \int \frac{1}{\sqrt{x}} \cdot \frac{1}{x+5} dx=\int \frac{1}{t^2+5} \cdot 2dt=2\int\frac{dt}{t^2+\sqrt{5}^2}=\frac{2}{\sqrt{5}} arctg \frac{t}{\sqrt{5}}+C=\frac{2}{\sqrt{5}} arctg \frac{\sqrt{x}}{\sqrt{5}}+C\)

\(\sqrt{x}=t \\
\frac{1}{2\sqrt{x}}dx=dt\\
x=t^2\)