Use a substitution to evaluate the given integral. ∫(x^2-1)...

Key Concepts

Algebraic Simplification

Algebraic simplification in the context of integration involves manipulating the integrand algebraically before or after substitution to reveal a standard integral form. This may include expanding expressions, combining like terms, or factoring, and it is essential for reducing an integral into a form that is more straightforward to integrate.

Integration of Composite Functions

When dealing with composite functions in integration, recognizing the inner function and its derivative is key to applying substitution effectively. By identifying these parts, one can transform the original integral into an expression involving only the new variable, facilitating the use of standard integration techniques.

U-Substitution

U-substitution is a method for evaluating integrals by substituting a portion of the integrand with a new variable, typically simplifying the integral into a more standard form. This technique effectively reverses the chain rule from differentiation, turning a composite function into a simpler one that is easier to integrate.

Change of Variables in Integration

The change of variables technique involves replacing the original variable with a new one to simplify an expression inside the integral. This method is especially useful when the integrand consists of a composition of functions, allowing the differential to be expressed in terms of the new variable, which often reduces the complexity of the integral.

Transcript

00:01 Okay, so before we try and actually compute the integral of this function, what we're going to do is we're going to get rid of this one -half -half in this 2x plus 1 to the 1 -half power by multiplying by 2x plus 1 to the 1 -half power divided by 2x plus 1 to the 1 -5 power.

00:15 So we're going to have in the numerator x squared plus or minus 1 multiplied by 2x plus 1.

00:22 And in our denominator, we're going to have 2x plus 1 to the 1 -half power.

00:27 And we also have dx here.

00:29 And now we're going to use integration by substitution, and we're going to let you here equal 2x plus 1.

00:36 So then du is equal to 2 times dx.

00:43 So then dx is equal to du divided by 2.

00:52 And so when we plug in du, we're going to have to divide by 2 as well.

00:57 And then so the next thing we want to notice is that if 2x plus 1 is equal to u, that means that u minus 1 divided by 2, squared is equal to x squared or u minus 1 divided by 2 is equal to x and so what we're going to do is we're going to plug these into our integral we're now going to have the integral of u minus 1 divided by 2 squared and it's going to be minus 1 and then we have u and u to the 1 half power and d u and we need to divide by 2 as well so i'm actually just going to put that one half out here.

01:40 And so what we're going to want to do is actually multiply or distribute this squared through this parentheses.

01:46 And so in the numerator, we're going to get u squared minus 2u plus one.

01:53 And in the denominator, we're going to get four.

01:56 And then we have minus one.

01:58 And this is all divided by u to the one -half power.

02:02 However, what we can do is we can just say that this first term is going to be divided by four times you to the one -half power and the second term is just divided by you.

02:12 And so what we want to do now is split up this first fraction into three different fractions.

02:19 So we're going to have the integral of u squared divided by four u to the one -half power minus two u divided by u to the one -half power and then plus one divided by four u.

02:33 To 1 .5 power and minus 1 divided by u multiplied by d u.

02:40 And so what we can do is we can minus the exponent in the denominator from that in the numerator...

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