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How do you integrate #x(x^2+1)^2 dx#?

Answer 1

Isabella Ackerman

#intx(x^2+1)^2dx=x^6/6+x^4/2+x^2/2+"C"#

Expand #(x^2+1)^2#

#(x^2+1)^2=x^4+2x^2+1#

Distribute the #x#

#x(x^4+2x^2+1)=x^5+2x^3+x#

Next we integrate each term

#intx^5+2x^3+xdx=intx^5dx+int2x^3dx+intxdx#

#=x^6/6+2*x^4/4+x^2/2#

#=x^6/6+x^4/2+x^2/2+"C"#

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Answer 2

Ezekiel Alexander

To integrate (x(x^2+1)^2 ,dx), you can use the substitution method. Let (u = x^2 + 1). Then, (du = 2x ,dx). Solving for (dx), we get (dx = \frac{du}{2x}). Substituting these into the integral, we have:

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

Now, integrate (u^2) with respect to (u), which gives:

[\frac{1}{2} \int u^2 ,du = \frac{1}{2} \cdot \frac{u^3}{3} + C = \frac{1}{6}u^3 + C]

Finally, substitute back (u = x^2 + 1):

[\frac{1}{6}(x^2 + 1)^3 + C]

So, the integral of (x(x^2+1)^2 ,dx) is (\frac{1}{6}(x^2 + 1)^3 + 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.