Integraali $$$\sqrt{a^{x} - 1}$$$:stä muuttujan $$$x$$$ suhteen

Määritä $$$\int \sqrt{a^{x} - 1}\, dx$$$.

Olkoon $$$u=\sqrt{a^{x} - 1}$$$.

Tällöin $$$du=\left(\sqrt{a^{x} - 1}\right)^{\prime }dx = \frac{a^{x} \ln{\left(a \right)}}{2 \sqrt{a^{x} - 1}} dx$$$ (vaiheet ovat nähtävissä ») ja saamme, että $$$\frac{a^{x} dx}{\sqrt{a^{x} - 1}} = \frac{2 du}{\ln{\left(a \right)}}$$$.

Näin ollen,

$${\color{red}{\int{\sqrt{a^{x} - 1} d x}}} = {\color{red}{\int{\frac{2 u^{2}}{\left(u^{2} + 1\right) \ln{\left(a \right)}} d u}}}$$

Sovella vakiokertoimen sääntöä $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ käyttäen $$$c=\frac{2}{\ln{\left(a \right)}}$$$ ja $$$f{\left(u \right)} = \frac{u^{2}}{u^{2} + 1}$$$:

$${\color{red}{\int{\frac{2 u^{2}}{\left(u^{2} + 1\right) \ln{\left(a \right)}} d u}}} = {\color{red}{\left(\frac{2 \int{\frac{u^{2}}{u^{2} + 1} d u}}{\ln{\left(a \right)}}\right)}}$$

Kirjoita murtolauseke uudelleen ja jaa se osamurtoihin:

$$\frac{2 {\color{red}{\int{\frac{u^{2}}{u^{2} + 1} d u}}}}{\ln{\left(a \right)}} = \frac{2 {\color{red}{\int{\left(1 - \frac{1}{u^{2} + 1}\right)d u}}}}{\ln{\left(a \right)}}$$

Integroi termi kerrallaan:

$$\frac{2 {\color{red}{\int{\left(1 - \frac{1}{u^{2} + 1}\right)d u}}}}{\ln{\left(a \right)}} = \frac{2 {\color{red}{\left(\int{1 d u} - \int{\frac{1}{u^{2} + 1} d u}\right)}}}{\ln{\left(a \right)}}$$

Sovella vakiosääntöä $$$\int c\, du = c u$$$ käyttäen $$$c=1$$$:

$$\frac{2 \left(- \int{\frac{1}{u^{2} + 1} d u} + {\color{red}{\int{1 d u}}}\right)}{\ln{\left(a \right)}} = \frac{2 \left(- \int{\frac{1}{u^{2} + 1} d u} + {\color{red}{u}}\right)}{\ln{\left(a \right)}}$$

Funktion $$$\frac{1}{u^{2} + 1}$$$ integraali on $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$:

$$\frac{2 \left(u - {\color{red}{\int{\frac{1}{u^{2} + 1} d u}}}\right)}{\ln{\left(a \right)}} = \frac{2 \left(u - {\color{red}{\operatorname{atan}{\left(u \right)}}}\right)}{\ln{\left(a \right)}}$$

Muista, että $$$u=\sqrt{a^{x} - 1}$$$:

$$\frac{2 \left(- \operatorname{atan}{\left({\color{red}{u}} \right)} + {\color{red}{u}}\right)}{\ln{\left(a \right)}} = \frac{2 \left(- \operatorname{atan}{\left({\color{red}{\sqrt{a^{x} - 1}}} \right)} + {\color{red}{\sqrt{a^{x} - 1}}}\right)}{\ln{\left(a \right)}}$$

Näin ollen,

$$\int{\sqrt{a^{x} - 1} d x} = \frac{2 \left(\sqrt{a^{x} - 1} - \operatorname{atan}{\left(\sqrt{a^{x} - 1} \right)}\right)}{\ln{\left(a \right)}}$$

Lisää integrointivakio:

$$\int{\sqrt{a^{x} - 1} d x} = \frac{2 \left(\sqrt{a^{x} - 1} - \operatorname{atan}{\left(\sqrt{a^{x} - 1} \right)}\right)}{\ln{\left(a \right)}}+C$$

$$$\int \sqrt{a^{x} - 1}\, dx = \frac{2 \left(\sqrt{a^{x} - 1} - \operatorname{atan}{\left(\sqrt{a^{x} - 1} \right)}\right)}{\ln\left(a\right)} + C$$$A