# Integral de $\sqrt{x^{2} + 1}$

La calculadora encontrará la integral/antiderivada de $\sqrt{x^{2} + 1}$, con los pasos que se muestran.

Escriba sin diferenciales como $dx$, $dy$ etc.
Deje vacío para la detección automática.

Si la calculadora no calculó algo o ha identificado un error, o tiene una sugerencia/comentario, escríbalo en los comentarios a continuación.

### Tu aportación

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

### Solución

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 $u=\operatorname{asinh}{\left(x \right)}$.

Therefore,

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

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

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

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

Thus,

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

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 can be rewritten as

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

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:

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

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