Integral of $$$\sqrt{x^{2} + 1}$$$

The calculator will find the integral/antiderivative of $$$\sqrt{x^{2} + 1}$$$, with steps shown.

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Find $$$\int \sqrt{x^{2} + 1}\, dx$$$.

Solution

Let $$$x=\sinh{\left(u \right)}$$$.

Then $$$dx=\left(\sinh{\left(u \right)}\right)^{\prime }du = \cosh{\left(u \right)} du$$$ (steps can be seen here).

Also, it follows that $$$u=\operatorname{asinh}{\left(x \right)}$$$.

Thus,

$$$\sqrt{x^{2} + 1} = \sqrt{\sinh^{2}{\left( u \right)} + 1}$$$

Use the identity $$$\sinh^{2}{\left( u \right)} + 1 = \cosh^{2}{\left( u \right)}$$$:

$$$\sqrt{\sinh^{2}{\left( u \right)} + 1}=\sqrt{\cosh^{2}{\left( u \right)}}$$$

$$$\sqrt{\cosh^{2}{\left( u \right)}} = \cosh{\left( u \right)}$$$

So,

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

Rewrite the hyperbolic cosine using the power reducing formula $$$\cosh^{2}{\left(\alpha \right)} = \frac{\cosh{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$ with $$$\alpha= u $$$:

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

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(u \right)} = \cosh{\left(2 u \right)} + 1$$$:

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

Integrate term by term:

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

Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:

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

Let $$$v=2 u$$$.

Then $$$dv=\left(2 u\right)^{\prime }du = 2 du$$$ (steps can be seen here), and we have that $$$du = \frac{dv}{2}$$$.

The integral becomes

$$\frac{u}{2} + \frac{{\color{red}{\int{\cosh{\left(2 u \right)} d u}}}}{2} = \frac{u}{2} + \frac{{\color{red}{\int{\frac{\cosh{\left(v \right)}}{2} d v}}}}{2}$$

Apply the constant multiple rule $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(v \right)} = \cosh{\left(v \right)}$$$:

$$\frac{u}{2} + \frac{{\color{red}{\int{\frac{\cosh{\left(v \right)}}{2} d v}}}}{2} = \frac{u}{2} + \frac{{\color{red}{\left(\frac{\int{\cosh{\left(v \right)} d v}}{2}\right)}}}{2}$$

The integral of the hyperbolic cosine is $$$\int{\cosh{\left(v \right)} d v} = \sinh{\left(v \right)}$$$:

$$\frac{u}{2} + \frac{{\color{red}{\int{\cosh{\left(v \right)} d v}}}}{4} = \frac{u}{2} + \frac{{\color{red}{\sinh{\left(v \right)}}}}{4}$$

Recall that $$$v=2 u$$$:

$$\frac{u}{2} + \frac{\sinh{\left({\color{red}{v}} \right)}}{4} = \frac{u}{2} + \frac{\sinh{\left({\color{red}{\left(2 u\right)}} \right)}}{4}$$

Recall that $$$u=\operatorname{asinh}{\left(x \right)}$$$:

$$\frac{\sinh{\left(2 {\color{red}{u}} \right)}}{4} + \frac{{\color{red}{u}}}{2} = \frac{\sinh{\left(2 {\color{red}{\operatorname{asinh}{\left(x \right)}}} \right)}}{4} + \frac{{\color{red}{\operatorname{asinh}{\left(x \right)}}}}{2}$$

Therefore,

$$\int{\sqrt{x^{2} + 1} d x} = \frac{\sinh{\left(2 \operatorname{asinh}{\left(x \right)} \right)}}{4} + \frac{\operatorname{asinh}{\left(x \right)}}{2}$$

Using the formulas $$$\sin{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\sin{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\cos{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 1 - 2 \alpha^{2}$$$, $$$\cos{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$, $$$\sinh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha^{2} + 1}$$$, $$$\sinh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha - 1} \sqrt{\alpha + 1}$$$, $$$\cosh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} + 1$$$, $$$\cosh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$, simplify the expression:

$$\int{\sqrt{x^{2} + 1} d x} = \frac{x \sqrt{x^{2} + 1}}{2} + \frac{\operatorname{asinh}{\left(x \right)}}{2}$$

Simplify further:

$$\int{\sqrt{x^{2} + 1} d x} = \frac{x \sqrt{x^{2} + 1} + \operatorname{asinh}{\left(x \right)}}{2}$$

Add the constant of integration:

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

Answer: $$$\int{\sqrt{x^{2} + 1} d x}=\frac{x \sqrt{x^{2} + 1} + \operatorname{asinh}{\left(x \right)}}{2}+C$$$