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3 years ago

AB passes through a A(-3,0) B(-6,5) what is the equation of the line that passes through the origin and is parallel to AB

1 answer:

6 0

Slope = (5 - 0)/(-6 + 3) = -5/3

<span>passes through the origin
</span>y = mx + b
0 = -5/3(-3) + b
b = -5

equation

y = -5/3x - 5

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Answer:

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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

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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:

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

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ExtremeBDS [4]

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