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How do you integrate #x5^(x^2) dx#?

Answer 1

Charles Anctil

#1/(2log_e5)5^(x^2)+C#

#d/(dx)(5^(x^2)) = 2 x log_e5 xx 5^(x^2)# then

#int x5^(x^2) dx = 1/(2log_e5)int d/(dx)(5^(x^2)) = 1/(2log_e5)5^(x^2)+C#

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Answer 2

Emilia Aguilar

To integrate (x^5^{x^2}) with respect to (x), you would typically use substitution. Let (u = x^2). Then, (du/dx = 2x), or (dx = du/(2x)). After substitution, you can integrate with respect to (u) and then convert back to the original variable (x).

The integral becomes:

[ \frac{1}{2} \int u^2 \cdot u^{\frac{1}{2}} , du ]

which simplifies to:

[ \frac{1}{2} \int u^{\frac{5}{2}} , du ]

Now integrate (\frac{1}{2} \int u^{\frac{5}{2}} , du) to get:

[ \frac{1}{2} \cdot \frac{2}{7} \cdot u^{\frac{7}{2}} + C ]

Substitute back for (u = x^2):

[ \frac{1}{7} x^{\frac{7}{2}} + 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.