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I have a question on how to evaluate this integral:

$\int (5-x^2+\frac{18}{x^4})dx$

Is this correct?

$\int (5-x^2+\frac{18}{x^4})dx$

$ \int 5 dx = 5x + C$

$ \int -x^2 = \frac {-x^3}{3}= -\frac{1}{3}x^3 +C $

$ \int \frac {18}{x^4} = \int 18x^{-4} dx = \frac {18x^{-3}}{3}=6x^{-3} + C$

So, $\int (5-x^2+\frac{18}{x^4})dx= 5x-\frac{1}{3}x^3+6x^{-3}+C$

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    Is it too much to ask you to use Mathjax? – 2017-02-01

3 Answers 3

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It is not quite correct. Note that

$$\int 18x^{-4}\,dx=18\,\left(\frac{x^{(-4+1)}}{(-4+1)}\right)+C=-6x^{-3}+C$$

On a side note, it is not good form to write "$5=5x$" as short hand notation for "$\int 5\,dx=5x$."

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    "*it is not good form to write...*" I completely agree. Equals signs should only be used for equalities. – 2017-02-01
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    Thank you both for telling me this (: – 2017-02-01
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    @Dan I recommend you write $5x\implies5$ to avoid confusion. – 2017-02-01
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    @JMoravitz While I stated the "more-than-obvious," it was clear, I believe, that the OP was not implying an equality. Rather, it was a shorthand notation that omitted the integral sign. And hence, this omission was the "bad form" reference. – 2017-02-01
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This is how you would do this

$$\int (5-x^2+\frac{18}{x^4})dx$$

Integration as you have probably shown spits across addition so we have;

$$ \int5dx+\int-x^2dx+\int\frac{18}{x^4}dx $$ now we can also take out constaints so we have:

$$ 5\int1dx-\int x^2dx+18\int x^{-4}dx $$ now by defintion we have $$ 5x-\frac{x^3}{3}+18\frac{x^{-3}}{-3}+C $$ Not we only need one C because three constants added together are constants $$ 5x-\frac{x^3}{3}-\frac{6}{x^3}+C $$

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Well we know what you mean but the equal sign is obviously wrong. You should use the integral of sum rule and use an integral sign before each of the summands. Also I think the question was regarding $dx$. That's just a notation to specify the variable you're integrating with respect to.