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docker41 [41]
3 years ago
7

What is the integer closest to the square root of 17

Mathematics
2 answers:
Alex Ar [27]3 years ago
7 0

Answer:

4

Step-by-step explanation:

\sqrt{16} = 4

faltersainse [42]3 years ago
6 0

Answer:

4.

Step-by-step explanation:

The square root of 17 is 4.12...

square root of 16 is 4

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Mandarinka [93]

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Evaluate the indefinite integral. <br> integar x4/1 + x^10 dx
ivann1987 [24]

Answer:

\int\ {\frac{x^4}{1 + x^{10}}} \, dx = \frac{1}{5}( \arctan(x^5)) + c

Step-by-step explanation:

Given

\int\ {\frac{x^4}{1 + x^{10}}} \, dx

Required

Integrate

We have:

\int\ {\frac{x^4}{1 + x^{10}}} \, dx

Let

u = x^5

Differentiate

\frac{du}{dx} = 5x^4

Make dx the subject

dx = \frac{du}{5x^4}

So, we have:

\int\ {\frac{x^4}{1 + x^{10}}} \, dx

\int\ {\frac{x^4}{1 + x^{10}}} \, \frac{du}{5x^4}

\frac{1}{5} \int\ {\frac{1}{1 + x^{10}}} \, du

Express x^(10) as x^(5*2)

\frac{1}{5} \int\ {\frac{1}{1 + x^{5*2}}} \, du

Rewrite as:

\frac{1}{5} \int\ {\frac{1}{1 + x^{5)^2}}} \, du

Recall that: u = x^5

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

Integrate

\frac{1}{5} * \arctan(u) + c

Substitute: u = x^5

\frac{1}{5} * \arctan(x^5) + c

Hence:

\int\ {\frac{x^4}{1 + x^{10}}} \, dx = \frac{1}{5}( \arctan(x^5)) + c

7 0
3 years ago
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