Evaluate the integral using any appropriate algebraic method or trigonometric identity. $$ \int \frac{3 \theta^{3}-7 \theta^{2}+13 \theta}{3 \theta-4} d \theta $$

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Calculate the integral $$ \int \frac{3 	heta^{3}-7 	heta^{2}+13 	heta}{3 	heta-4} d 	heta $$.
Evaluate the integral by following steps:
- step0: Evaluate using formulas and rules:
  $$\int \frac{3	heta ^{3}-7	heta ^{2}+13	heta }{3	heta -4} d	heta$$
- step1: Rewrite the fraction:
  $$\int \left(	heta ^{2}-	heta +3+\frac{12}{3	heta -4}ight) d	heta$$
- step2: Use properties of integrals:
  $$\int \left(	heta ^{2}-	heta +3ight) d	heta+\int \frac{12}{3	heta -4} d	heta$$
- step3: Evaluate the integral:
  $$\frac{	heta ^{3}}{3}-\frac{	heta ^{2}}{2}+3	heta +\int \frac{12}{3	heta -4} d	heta$$
- step4: Evaluate the integral:
  $$\frac{	heta ^{3}}{3}-\frac{	heta ^{2}}{2}+3	heta +4\ln{\left(\left|3	heta -4ight|ight)}$$
- step5: Add the constant of integral C:
  $$\frac{	heta ^{3}}{3}-\frac{	heta ^{2}}{2}+3	heta +4\ln{\left(\left|3	heta -4ight|ight)} + C, C \in \mathbb{R}$$
The integral of the given function is $$\frac{	heta^{3}}{3}-\frac{	heta^{2}}{2}+3	heta +4\ln{(3	heta -4)} + C$$, where $$C$$ is an arbitrary constant.
The integral evaluates to $$\frac{	heta^{3}}{3}-\frac{	heta^{2}}{2}+3	heta +4\ln{(3	heta -4)} + C$$.

Evaluate the integral using any appropriate algebraic method or trigonometric identity.

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