1. how to integrate (secinverse sqrtx – taninverse sqrtx) /(secinverse sqrtx + taninverse sqrtx)

1. how to integrate e^x(x^5 – x + 2 )/(x^2+1)^2

To tackle the integration of the expressions you've provided, we can break them down into manageable parts. Let's start with the first expression: integrating \((\sec^{-1}(\sqrt{x}) - \tan^{-1}(\sqrt{x})) / (\sec^{-1}(\sqrt{x}) + \tan^{-1}(\sqrt{x}))\).

Integration of the First Expression

To integrate this expression, we can use substitution and properties of inverse trigonometric functions. Let's denote:

• u = \sec^{-1}(\sqrt{x})
• v = \tan^{-1}(\sqrt{x})

Thus, we can rewrite the integral as:

∫ (u - v) / (u + v) du

Next, we can simplify the integrand. Notice that:

u + v = \sec^{-1}(\sqrt{x}) + \tan^{-1}(\sqrt{x})

To integrate this, we can use the identity for the derivative of inverse functions. The derivatives of these functions can be calculated, and we can express the integral in terms of x. However, this integral can become complex, and often, numerical methods or software tools are used for evaluation.

Integration of the Second Expression

Now, let's move on to the second expression: integrating \(\frac{e^x(x^5 - x + 2)}{(x^2 + 1)^2}\).

This integral can be approached using integration by parts or recognizing patterns in the numerator and denominator. A useful strategy here is to apply integration by parts, where we let:

• u = e^x
• dv = \frac{x^5 - x + 2}{(x^2 + 1)^2} dx

Then, we differentiate and integrate accordingly:

• du = e^x dx
• v = ∫ \frac{x^5 - x + 2}{(x^2 + 1)^2} dx

Now, the integral can be expressed as:

∫ u dv = uv - ∫ v du

Calculating \(v\) requires breaking down the fraction \(\frac{x^5 - x + 2}{(x^2 + 1)^2}\) using polynomial long division or partial fractions, depending on the complexity. This will allow us to integrate \(v\) more easily.

Final Steps

After finding \(v\), substitute back into the integration by parts formula. The final result will involve combining the terms and simplifying. If you encounter any specific difficulties with the integration steps, feel free to ask for clarification on those points!

In summary, both integrals require careful manipulation and sometimes the application of integration techniques like substitution or integration by parts. Each step builds on the previous one, leading to a solution that can be expressed in terms of elementary functions or evaluated numerically if necessary.