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[tex]I=\int \sqrt{(tanh(x))}[/tex]

La $u = \sqrt{\tanh(x)}$, slik at ${\mathrm d}u = \dfrac{(1-\tanh^2(x))}{2\sqrt{\tanh(x)}}{\mathrm d}x = \dfrac{1-u^4}{2u}{\mathrm d}x$. Dette gir
$$\begin{align*}
I &= \int u\left(\dfrac{2u}{1-u^4}{\mathrm d}u\right)\\
&= \int\dfrac{2u^2}{1-u^4}{\mathrm d}u\\
&= -\int\dfrac{(u^2 + 1) + (u^2 - 1)}{u^4-1}{\mathrm d}u\\
&= -\int\dfrac{1}{u^2-1}{\mathrm d}u-\int\dfrac{1}{u^2+1}{\mathrm d}u\\
\end{align*}$$
Her er $$\int\dfrac{1}{u^2-1}{\mathrm d}u = \dfrac12\int\dfrac{u+1-(u-1)}{u^2-1}{\mathrm d}u = \dfrac12\int\left(\dfrac{1}{u-1}-\dfrac{1}{u+1}\right){\mathrm d}u = \dfrac12\ln\left|\dfrac{u-1}{u+1}\right|$$
og $$\int\dfrac{1}{u^2+1}{\mathrm d}u = \arctan(u).$$
Dermed er $$I = \dfrac12 \ln\left|\dfrac{u+1}{u-1}\right| - \arctan(u) = \dfrac12 \ln\left|\dfrac{\sqrt{\tanh(x)}+1}{\sqrt{\tanh(x)}-1}\right| - \arctan\left(\sqrt{\tanh(x)}\right)+C$$

MatIsa skrev:La $u = \sqrt{\tanh(x)}$, slik at ${\mathrm d}u = \dfrac{(1-\tanh^2(x))}{2\sqrt{\tanh(x)}}{\mathrm d}x = \dfrac{1-u^4}{2u}{\mathrm d}x$. Dette gir
$$\begin{align*}
I &= \int u\left(\dfrac{2u}{1-u^4}{\mathrm d}u\right)\\
&= \int\dfrac{2u^2}{1-u^4}{\mathrm d}u\\
&= -\int\dfrac{(u^2 + 1) + (u^2 - 1)}{u^4-1}{\mathrm d}u\\
&= -\int\dfrac{1}{u^2-1}{\mathrm d}u-\int\dfrac{1}{u^2+1}{\mathrm d}u\\
\end{align*}$$
Her er $$\int\dfrac{1}{u^2-1}{\mathrm d}u = \dfrac12\int\dfrac{u+1-(u-1)}{u^2-1}{\mathrm d}u = \dfrac12\int\left(\dfrac{1}{u-1}-\dfrac{1}{u+1}\right){\mathrm d}u = \dfrac12\ln\left|\dfrac{u-1}{u+1}\right|$$
og $$\int\dfrac{1}{u^2+1}{\mathrm d}u = \arctan(u).$$
Dermed er $$I = \dfrac12 \ln\left|\dfrac{u+1}{u-1}\right| - \arctan(u) = \dfrac12 \ln\left|\dfrac{\sqrt{\tanh(x)}+1}{\sqrt{\tanh(x)}-1}\right| - \arctan\left(\sqrt{\tanh(x)}\right)+C$$

Såvidt jeg ser, ser det fint ut det der. Kjekk løsning.

Et alternativ er [tex]\frac{2u^2}{1-u^4}=\frac{1}{1-u^2}-\frac{1}{1+u^2}[/tex] slik at vi kunne ha sagt at [tex]\int \frac{1}{1-u^2}=arctanh(u)[/tex] og [tex]\int \frac{1}{1+u^2}=arctan(u)[/tex]

Som i alt gir [tex]tanh^{-1}(\sqrt{tanh(x)})-tan^{-1}(\sqrt{tanh(x)})+C[/tex]