Integrate( 1\x^3-3x-4)dx

Ответ:

karenrivera22

08.02.2022

$-\frac{1}{2x^2} - \frac{3}{2}x^2 -4x +c$ Assuming you need the integral this would be the answer.

Step-by-step explanation:

$\int\limits \frac{1}{x^3} -3x \, -4dx$

Thats after removing parenthesis.

Now we have to split into multiple intergrals..

$\int\ \frac{1}{x^3} \, dx + \int\ -3xdx + \int\-4dx$

Now we want to move x^3 out of the denom by raising it to the -1 power

$\int\ (x^3)^{-1}dx + \int\- -3xdx+\int\ -4dx$

Multiply the exponents in (x^3)^-1

We also need to apply the power rule of (a^m)^n = a^mn

$\int\ x^{3*-1}dx+\int\ -3xdx + \int\ -4dx$

3 by -1

$\int\ x^{-3}dx+\int\ -3xdx + \int\ -4dx$

Now we want to use the power rule (I assume you know this rule so I won't be posting it) as I with the integral of x^-3 where x is -1/2x^-2

$-\frac{1}{2}x^{-2} + C \int\ -3xdx + \int\ - 4dx$

Now since -3 is a constant we can move -3 out of the integral since it respects x.

$-\frac{1}{2}x^{-2} + C - 3 \int\ xdx + \int\ -4dx$

Now we need to remember the power rule with the respect to 1/2x^2 which is respective to x.

$-\frac{1}{2}x^{-2} + C - 3 (\frac{1}{2}x^2 +C)+\int -4dx$

Then we apply the constat rule (You should know this as well)

$-\frac{1}{2}x^{-2} + C - 3 (\frac{1}{2}x^2+C)-4x+C$

Now we simplify....

$-\frac{1}{2}x^{-2} + C - 3 (\frac{x^2}{2}+C)-4x + C -\frac{1}{2x^2} - \frac{3x^2}{2} -4x + C$

As you can see you just then re-order as above.

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