How do you use substitution to integrate #(2x(x^2 + 1)^23)dx#?

Answer 1

#int[2x(x^2+1)^23]dx=1/24(x^2+1)^24+C#

Use the substitution : Let #u=x^2+1#
Then #du=2xdx# so this can be plugged into the integral. It may help to rewrite it so #2x# and #dx# are next to one another:
#int(x^2+1)^23(2xdx)#

We may hence write the original integral in terms of x into a new equivalent integral in terms of u as follows :

#intu^23du=u^24/24+C#
We now substitute back in terms of x to obtain the final answer as #(x^2+1)^24/24+C#
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Answer 2

To integrate (2x(x^2 + 1)^{23}) using substitution, let (u = x^2 + 1). Then, (du = 2x , dx).

Substitute (u) and (du) into the integral:

[\int 2x(x^2 + 1)^{23} , dx = \int u^{23} , du]

Now integrate with respect to (u):

[\int u^{23} , du = \frac{u^{24}}{24} + C]

Replace (u) with (x^2 + 1) and simplify:

[\frac{(x^2 + 1)^{24}}{24} + C]

Therefore, the integral of (2x(x^2 + 1)^{23}) is (\frac{(x^2 + 1)^{24}}{24} + C), where (C) is the constant of integration.

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Answer from HIX Tutor

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.

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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