evaluate the intergral using substitution
ingergral[(x^2-e^3x)/(x^3-e^3x)^2]dx
By the "Just Look-at-it Theorem and Knowing your Derivatives" and knowing the derivative of ln(...)
I noticed that the derivative of the bottom would be
3(x^2 - e^3x) , which just happens to look awfully close to the top
so if y' = [(x^2-e^3x)/(x^3-e^3x)^2] then
y = (1/3)(ln(x^3 - e^3x))
To evaluate the given integral using substitution, we can make the substitution \(u = x^3 - e^{3x}\). Then we can find \(du\) in terms of \(dx\) to rewrite the integral in terms of \(u\) and calculate it.
Let's go through the steps in detail:
Step 1: Make the substitution
Let \(u = x^3 - e^{3x}\). Taking the derivative of \(u\) with respect to \(x\), we get:
\(du = (3x^2 - 3e^{3x})dx\)
Step 2: Rewrite the integral with the substitution variables
Next, we can substitute the expression for \(u\) and \(du\) into the original integral to obtain:
\(\int\frac{{x^2 - e^{3x}}}{{(x^3- e^{3x})^2}}dx = \int\frac{1}{{u^2}}du\)
Step 3: Evaluate the integral in terms of the new variable
The integral \(\int\frac{1}{{u^2}}du\) can be easily evaluated as:
\(-\frac{1}{u} + C\), where \(C\) is the constant of integration.
Step 4: Substitute back the original variable
Finally, substitute back the original variable \(x\) for \(u\) to get the answer in terms of \(x\):
\(-\frac{1}{{x^3 - e^{3x}}} + C\), where \(C\) is the constant of integration.
Therefore, the evaluated integral is \(-\frac{1}{{x^3 - e^{3x}}} + C\).