$$ \int \frac{ ext { Factor the diferonce of two squaros. }}{36-x^{2}} 33-x^{2}=\square $$

Question

$$ \int \frac{	ext { Factor the diferonce of two squaros. }}{36-x^{2}} 33-x^{2}=\square $$

UpStudy AI · Accepted Answer

Calculate or simplify the expression $$ \int \frac{33-x^{2}}{36-x^{2}} $$.
Evaluate the integral by following steps:
- step0: Evaluate using formulas and rules:
  $$\int \frac{33-x^{2}}{36-x^{2}} dx$$
- step1: Rearrange the terms:
  $$\int -\frac{33-x^{2}}{x^{2}-36} dx$$
- step2: Rewrite the fraction:
  $$\int -\left(-1-\frac{3}{x^{2}-36}\right) dx$$
- step3: Calculate:
  $$\int \left(1+\frac{3}{x^{2}-36}\right) dx$$
- step4: Use properties of integrals:
  $$\int 1 dx+\int \frac{3}{x^{2}-36} dx$$
- step5: Evaluate the integral:
  $$x+\int \frac{3}{x^{2}-36} dx$$
- step6: Evaluate the integral:
  $$x+\frac{1}{4}\ln{\left(\left|x-6\right|\right)}-\frac{1}{4}\ln{\left(\left|x+6\right|\right)}$$
- step7: Subtract the terms:
  $$x+\frac{1}{4}\ln{\left(\frac{\left|x-6\right|}{\left|x+6\right|}\right)}$$
- step8: Add the constant of integral C:
  $$x+\frac{1}{4}\ln{\left(\frac{\left|x-6\right|}{\left|x+6\right|}\right)} + C, C \in \mathbb{R}$$
The integral of $$ \frac{33-x^{2}}{36-x^{2}} $$ with respect to $$ x $$ is given by:

$$ \int \frac{33-x^{2}}{36-x^{2}} dx = x + \frac{1}{4} \ln\left(\frac{|x-6|}{|x+6|}\right) + C $$

where $$ C $$ is the constant of integration.
The integral of $$ \frac{33 - x^{2}}{36 - x^{2}} $$ with respect to $$ x $$ is: $$ x + \frac{1}{4} \ln\left(\frac{|x-6|}{|x+6|}\right) + C $$ where $$ C $$ is the constant of integration.

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