# How do you integrate #int (x^2+1)sqrt(x-2)# using substitution?

the obvious thing is to simplify that radical so that it can be distributed

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To integrate ( \int (x^2 + 1)\sqrt{x - 2} ) using substitution, let's make the substitution ( u = x - 2 ). Then, we have ( du = dx ).

Substitute ( u = x - 2 ) into the expression:

[ \int (x^2 + 1)\sqrt{x - 2} , dx = \int ((u + 2)^2 + 1)\sqrt{u} , du ]

Expand ( (u + 2)^2 ) and simplify:

[ \int (u^2 + 4u + 4 + 1)\sqrt{u} , du = \int (u^2 + 4u + 5)\sqrt{u} , du ]

Now, distribute the square root:

[ \int (u^2\sqrt{u} + 4u\sqrt{u} + 5\sqrt{u}) , du ]

Now, integrate each term separately:

[ = \frac{u^{\frac{5}{2}}}{\frac{5}{2}} + 4\frac{u^{\frac{3}{2}}}{\frac{3}{2}} + 5\frac{u^{\frac{1}{2}}}{\frac{1}{2}} + C ]

[ = \frac{2}{5}u^{\frac{5}{2}} + \frac{8}{3}u^{\frac{3}{2}} + 10u^{\frac{1}{2}} + C ]

Finally, substitute back ( u = x - 2 ) to get the final result:

[ = \frac{2}{5}(x - 2)^{\frac{5}{2}} + \frac{8}{3}(x - 2)^{\frac{3}{2}} + 10(x - 2)^{\frac{1}{2}} + C ]

This is the integral of ( (x^2 + 1)\sqrt{x - 2} ) using substitution.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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