Question

Integrate x^2/4-x^3​dx

Answer 1

Answer:

$\frac{-1}{3} \ln|4-x^3|+C$ is the answer

if $\int \frac{x^2 dx}{4-x^3}$ or $\int \frac{x^2}{4-x^3} dx$ was the integral.

Step-by-step explanation:

$\int \frac{x^2 dx}{4-x^3}$.

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:

$\frac{du}{dx}=0-3x^2$

$du=-3x^2 dx$

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

Divide both sides by -3:

$\frac{-1}{3}du=x^2 dx$

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

$\int \frac{x^2 }{4-x^3}dx=int \frac{x^2 dx}{4-x^3}$

with $u=4-x^3$ and $\frac{-1}{3}du=x^2 dx$:

$\int \frac{-1}{3} du}{u}$

$\frac{-1}{3} \int \frac{du}{u}$

$\frac{-1}{3} \int u^{-1} du$

Don't use power rule.

$\frac{-1}{3} \ln|u|+C$

Put back in terms of x:

$\frac{-1}{3} \ln|4-x^3|+C$

Let's check our answer by differentiating it:

$\frac{-1}{3}(\ln|4-x^3|)'$

$\frac{-1}{3} \cdot \frac{-3x^2}{4-x^3}$

$\frac{x^2}{4-x^3}$

We are good. Our check it worked out.