2 years ago

(50 points and Brainiest

1 answer:

7 0

could you possibly make two zoomed in pictures?

its quite hard to see

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olga nikolaevna [1]

Simplify the integrand:

$\dfrac{4x^4 + 3x^2 + 5x}{x^2} = 4x^2 + 3 + \dfrac5x$

Now integrate one term at a time.

$\displaystyle \int 4x^2 \, dx = \frac43 x^3 + C$

$\displaystyle \int 3 \, dx = 3x + C$

$\displaystyle \int \frac5x \, dx = 5 \ln|x| + C$

Putting everything together,

$\displaystyle \int \frac{4x^4 + 3x^2 + 5x}{x^2} \, dx = \int \left(4x^2 + 3 + \frac5x\right) \, dx = \boxed{\frac43 x^3 + 3x + 5 \ln|x| + C}$

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1 year ago

kaheart [24]

Answer:

You started with 150.

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2 years ago

Zepler [3.9K]

Answer: A, B, and C are positive constants and that x+y= C. Show that the minimum value of +Ax%5E2%2BBy%5E2+ occurs when .

Step-by-step explanation:

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2 years ago

icang [17]

We use letters to substitute for unknown numbers. I hope this helps. :)

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3 years ago

MariettaO [177]

Answer:122

Step-by-step explanation:

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2 years ago