What is the integral of arcsinh(x) dx?

What is the integral of arcsinh(x) dx?

• A. x*arcsinh(x) + sqrt(x^2 + 1) + C
• B. x*arcsinh(x) - sqrt(x^2 + 1) + C
• C. x*arcsinh(x) + sqrt(1 + x^2) + C
• D. ln|x + sqrt(x^2 + 1)| + C

Answer: (B)

To determine the integral of arcsinh(x) dx, it is helpful to understand the antiderivative of the inverse hyperbolic sine function. The integral of arcsinh(x) can be approached using integration by parts. Letting \( u = \text{arcsinh}(x) \) and \( dv = dx \), we have: - \( du = \frac{1}{\sqrt{x^2 + 1}} dx \) - \( v = x \) Applying the integration by parts formula, \( \int u \, dv = uv - \int v \, du \), we have: \[ \int \text{arcsinh}(x) \, dx = x \cdot \text{arcsinh}(x) - \int x \cdot \frac{1}{\sqrt{x^2 + 1}} \, dx \] The remaining integral, \( \int x \cdot \frac{1}{\sqrt{x^2 + 1}} \, dx \), can be solved by using the substitution \( u = x^2 + 1 \), which simplifies the integral to \( \int \frac{1}{2} u^{-1/

To determine the integral of arcsinh(x) dx, it is helpful to understand the antiderivative of the inverse hyperbolic sine function.

The integral of arcsinh(x) can be approached using integration by parts. Letting ( u = \text{arcsinh}(x) ) and ( dv = dx ), we have:

• ( du = \frac{1}{\sqrt{x^2 + 1}} dx )

• ( v = x )

Applying the integration by parts formula, ( \int u , dv = uv - \int v , du ), we have:

[

\int \text{arcsinh}(x) , dx = x \cdot \text{arcsinh}(x) - \int x \cdot \frac{1}{\sqrt{x^2 + 1}} , dx

]

The remaining integral, ( \int x \cdot \frac{1}{\sqrt{x^2 + 1}} , dx ), can be solved by using the substitution ( u = x^2 + 1 ), which simplifies the integral to ( \int \frac{1}{2} u^{-1/