# integration of $\sqrt{x^{2}+1}$

The integral

 $I\,:=\,\int\!\sqrt{x^{2}\!+\!1}\;dx$

can be found by using the first Eulerβs substitution (http://planetmath.org/EulersSubstitutionsForIntegration)

 $\sqrt{x^{2}\!+\!1}\;:=\;-x\!+\!t,$

but another possibility is to use partial integration (http://planetmath.org/ASpecialCaseOfPartialIntegration) if one knows the integral $\int\!\frac{dx}{\sqrt{x^{2}+1}}$.β The corresponding may be said of the more general

 $\int\!\sqrt{x^{2}\!+\!c}\;dx.$

We think that the integrand of $I$ has the other factor 1 and integrate partially:

 $I\,=\,\int\!1\cdot\sqrt{x^{2}\!+\!1}\;dx\,=\,x\sqrt{x^{2}\!+\!1}-\!\int\!x% \cdot\frac{1}{2\sqrt{x^{2}\!+\!1}}\cdot 2x\,dx+C^{\prime}\,=\,x\sqrt{x^{2}\!+% \!1}-\!\int\!\frac{x^{2}}{\sqrt{x^{2}\!+\!1}}\,dx+C^{\prime}.$

Writing the numerator as $(x^{2}\!+\!1)\!-\!1$ and dividing its minuend and subtrahend separately, we can write

 $I\,=\,x\sqrt{x^{2}\!+\!1}-\!\left(\!\int\!\sqrt{x^{2}\!+\!1}\,dx-\!\int\!\frac% {1}{\sqrt{x^{2}\!+\!1}}\,dx\right)+C^{\prime}\,=\,x\sqrt{x^{2}\!+\!1}-I+\!\int% \!\frac{dx}{\sqrt{x^{2}\!+\!1}}+C^{\prime}.$

Having $I$ in two , we solve it from these equalities, obtaining

 $I\,=\,\frac{x}{2}\sqrt{x^{2}\!+\!1}+\frac{1}{2}\!\int\frac{dx}{\sqrt{x^{2}\!+% \!1}}+C,$

i.e.,

 $\int\!\sqrt{x^{2}\!+\!1}\;dx\;=\;\frac{x}{2}\sqrt{x^{2}\!+\!1}+\frac{1}{2}% \operatorname{arsinh}{x}+C$
Title integration of $\sqrt{x^{2}+1}$ IntegrationOfsqrtx21 2013-03-22 18:06:58 2013-03-22 18:06:58 pahio (2872) pahio (2872) 7 pahio (2872) Derivation msc 26A36 antiderivative of $\sqrt{x^{2}+1}$ IntegrationByParts AreaFunctions DerivativeOfInverseFunction