72.6k views
2 votes
Integrate x^2/4-x^3​dx

1 Answer

1 vote

Answer:


(-1)/(3) \ln|4-x^3|+C is the answer

if
\int (x^2 dx)/(4-x^3) or
\int (x^2)/(4-x^3) dx was the integral.

Explanation:


\int (x^2 dx)/(4-x^3).

I know the derivative of
x^3 will give me
3x^2 and I see the variable part of this in the numerator.

So my subsitution will be
u=4-x^3 and differentiating both sides gives:


(du)/(dx)=0-3x^2


du=-3x^2 dx

I'm going to solve for
x^2 dx since that is my numerator.

Divide both sides by -3:


(-1)/(3)du=x^2 dx

Inputting my substitution with it's derivative into the integral gives me:


\int (x^2 )/(4-x^3)dx=int (x^2 dx)/(4-x^3)

with
u=4-x^3 and
(-1)/(3)du=x^2 dx:


\int (-1)/(3) du}{u}


(-1)/(3) \int (du)/(u)


(-1)/(3) \int u^(-1) du

Don't use power rule.


(-1)/(3) \ln|u|+C

Put back in terms of x:


(-1)/(3) \ln|4-x^3|+C

Let's check our answer by differentiating it:


(-1)/(3)(\ln|4-x^3|)'


(-1)/(3) \cdot (-3x^2)/(4-x^3)


(x^2)/(4-x^3)

We are good. Our check it worked out.

User Arvin
by
7.2k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories