In Exercises 27–40, use a substitution to change the integral ...

Question

In Exercises 27–40, use a substitution to change the integral into one you can find in the table. Then evaluate the integral.

∫ (x^2 + 6x) / (x^2 + 3)^2 dx

Accepted Answer

Hello. In this video we are going to be evaluating the following integral. We are given the integral of the quantity x2 + 4 x divided by the quantity x2 + 5, all raised to the power of 2. Now, we can go ahead and algebraically simplify this integral by separating it into a sum of integrals. We can rewrite the integral to be the integral of x2 divided by X2 + 5 quantity 2, plus the integral of 4 x divided by the quantity x2 + 5 quantity squad X. We are going to label the first integral as 1 and the second integral as 2, and we are going to independently evaluate each integral first. Now, let's go ahead and start with the 2nd integral, because it's the easiest one to work with in comparison to the other one. Now if we're taking a look at the 2nd integral, we have a constant of 4 in the numerator. If we were to pull out that 4 to the front of the integral, we will rewrite the integral as 4 multiply the integral of X divided by the quantity X2 + 5, quantity squared DX. Now here we can apply a u substitution. Let u equal to x 2 plus 5. Then d u is equal to 2 x dx, which further simplifies to be 1/2 duu equal to x dx. By applying these substitutions, we get 2 multiplied by the integral of 1 divided by u squared duu. Now 1 divided by U 2 is equivalent to u to the power of -2. And so this is going to integrate to be 2 multiplied by u to the power of -1, which is equivalent to -2 divided by you. And by resubstituting our value of u, we get -2 divided by x 2 + 5. So this is going to be the evaluated form of the second integral. Now let's go ahead and evaluate the first integral. Now, the first integral is the integral of X2 divided by X2 + 5, raised the power of 2. Now here what we can go ahead and do is we can cleverly simplify this denominator. By introducing a sum and difference of terms of 5. So for example, let's go ahead and introduce a 5-5 to the numerator. Now the reason why we do this is because we can simplify this easier by separating this into a difference of integrals. If we were to separate this integral into two different integrals, we get the integral of X2 + 5 divided by X2 + 5 quantity 2 minus the integral of 5 divided by X2 + 5 quantity square. Now let's go ahead and evaluate this first integral. Now this integral is going to simplify to be the integral of 1 divided by x2 + 5. And this is going to evaluate to be the natural logarithm. No, I'm sorry, this is the identity for tangent inverse. So this is going to integrate to be 1 divided by the square root of 5 multiplied by tangent inverse of X divided by the square root of 5. Now let's take a look at the 2nd integral. Now, the 2nd integral is -5 multiplied by the integral of 1 divided by X2 + 5 quantity squared. Now this integral is going to simplify. To be X We can apply an identity to this integral. Now the identity is the following. If we have an integral of the form one divided. By x plus a quantity squared. All raises the power of 2. Then this is equivalent to X divided by 2A2 multiplied by X2 + A2. Plus 1 divided by 2 a cubed multiplied by the tangent inverse of x divided by a. Now here in this case a is going to equal to the square root of 5, so this is going to evaluate to be. X divided By 10 multiplied by x2 + 5 + 1 divided by 10 square root of 5 multiplied by the tangent inverse of x divided by the square root of 5. So this is just the identity for 1 divided by X2 + 5 quantity squared without distributing the -5. But if we were to distribute -5, so now we multiply this entire thing by -5. This we evaluate to be negative x divided by 2 multiplied by x2 + 5 minus 1 divided by 2 square root of 5 multiplied by the tangent inverse of x divided by the square root of 5. Now, if we were to combine these two parts of the of the integral together, by combining like terms, the first integral is going to simplify to be 1 divided by the square root of 5, tangent inverse of X divided by the square root of 5. Minus x divided by 2 multiplied by x2 + 5. So this is the first integral. Now keep in mind that earlier we saw for the second integral, which is -2 divided by x2 + 5. So combining both of the results together gives us 1 divided by the square root of 5 multiplied by the tangent inverse of x divided by the square root of 5. Minus x divided by 2 multiplied by x2 + 5. Plus. No, -2 divided by X2 + 5. And by combining like terms together, this integram is going to fully simplify to be 1 divided by the square root of 5 multiplied by the tangent inverse of x divided by the square root of 5. Minus X + 4 divided by 2 X2 + 10 + C. This is going to be the final solution to the given integral, and with that being said, the overall solution to the problem is going to be a. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

In Exercises 27–40, use a substitution to change the integral into one you can find in the table. Then evaluate the integral.∫ (x^2 + 6x) / (x^2 + 3)^2 dx

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In Exercises 27–40, use a substitution to change the integral into one you can find in the table. Then evaluate the integral.
∫ (x^2 + 6x) / (x^2 + 3)^2 dx