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How do you integrate #x(x²+1)³ dx #?

Answer 1

Evelyn Agard

We have that

#int x(x^2+1)^3dx=int 1/2*(x^2+1)'*(x^2+1)^3dx=1/8*(x^2+1)^4+c#

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Answer 2

Emilia Aguilar

To integrate ( x(x^2 + 1)^3 ) with respect to ( x ), you can use substitution. Let ( u = x^2 + 1 ). Then, ( du/dx = 2x ) or ( dx = du / (2x) ). Substitute these into the integral:

[ \begin{align*} \int x(x^2 + 1)^3 dx &= \int x u^3 \frac{du}{2x} \ &= \frac{1}{2} \int u^3 du \ &= \frac{1}{2} \left( \frac{u^4}{4} + C \right) \ &= \frac{1}{8} (x^2 + 1)^4 + C \end{align*} ]

So, the integral of ( x(x^2 + 1)^3 ) with respect to ( x ) is ( \frac{1}{8} (x^2 + 1)^4 + 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.