Question

Use formal substitution (as illustrated in Examples 5 and 6 ) to find the indefinite integral. $$ \int \frac{x}{\sqrt{x^{2}+25}} d x $$

Short answer

The indefinite integral of the function \( \frac{x}{\sqrt{x^{2}+25}} \) is \( (x^{2}+25)^{1/2} + C \).

Step-by-step solution

Identify and define the substitution variable

Let \( u = x^{2}+25 \). This is the most complicated part of the function and substituting it will simplify the integral. Take the differential of \(u\), \( du = 2xdx \). Calculate \( dx \) from this, \( dx = \frac{du}{2x} \).

Substitute \( u \) and \( dx \) in the integral

Replace \( x^{2}+25 \) with \( u \) and \( dx \) with \( \frac{du}{2x} \) giving: \( \int \frac{x}{\sqrt{u}}\times\frac{du}{2x} = \frac{1}{2}\int u^{-1/2} du \).

Evaluate the integral

Now has become a simple power rule integration problem: which can be solved as: \( \frac{1}{2}\int u^{-1/2} du = u^{1/2} + C \), where \(C\) is the constant of integration.

Substitute the original variable back

Substitute \( u = x^{2}+25 \) back into the integral: \( u^{1/2} + C = (x^{2}+25)^{1/2}+ C \).

Simplify if possible

The expression can't be further simplified, so, the answer to the integral is \( (x^{2}+25)^{1/2} + C \).

Key concepts

Formal Substitution

Formal substitution is a method used to simplify complex integrals. It's particularly useful when faced with complicated functions within an integral. By substituting a part of the expression with a new variable, we can transform the integral into a simpler form.

Here's how it works in the context of the given problem:

• Identify a section of the integral that can be replaced. This is often the most complicated part, such as a nested function or radical expression. For example, in the problem, we identify the part \( x^2 + 25 \) and choose \( u = x^2 + 25 \).
• Calculate the differential of \( u \), which is \( du = 2x\,dx \). This step helps in substituting \( dx \) later.
• Solve for \( dx \) in terms of \( du \), for this integral \( dx = \frac{du}{2x} \). This allows us to replace \( dx \) in the original integral.

With this, the original integral becomes more manageable. It transforms from a complex form into one that can be tackled with simpler integration techniques.

Power Rule Integration

The power rule is a simple yet powerful tool used in calculus for finding definite and indefinite integrals of functions in the form \( x^n \). For integrals, the power rule states that to integrate \( x^n \), you simply increase the exponent by 1 and then divide by the new exponent, except when \( n = -1 \).

This case becomes:

• \( \int x^n dx = \frac{x^{n+1}}{n+1} + C \) where \( n eq -1 \).

For the substitution example, after simplifying, the integral of \( u^{-1/2} \) is evaluated as follows:

• Apply the power rule: Increase the exponent by 1, turning \( -1/2 \) into \( 1/2 \).
• Then divide by the new exponent: \( \frac{u^{1/2}}{1/2} \) simplifies to \( 2u^{1/2} \).

Thus, when using power rule integration combined with substitution, the seemingly complex process becomes straightforward, giving us results that can be used to simplify the computation of the original integral.

Differential Calculus

Differential calculus is fundamental in mathematics, dealing with the concept of a derivative and how functions change. It allows us to find the rate at which quantities change and is a crucial building block of integral calculus.

In the context of our exercise, differential calculus is used to find the derivative of an expression during substitution. In this integration problem, after substituting \( u = x^2 + 25 \), you compute the differential \( du \) as follows:

• The derivative of \( x^2 + 25 \) with respect to \( x \) is \( 2x \).
• Thus, \( du = 2x\,dx \) encapsulates how the change in \( u \) depends on changes in \( x \).

This step is pivotal because it connects the new variable, \( u \), back to the original variable and helps substitute the differential \( dx \). Such processes showcase the tight link between differential calculus and integral calculus, where differentiation helps simplify integration.